板子们
单调队列——滑动窗口
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e6 + 3;
int n, k, a[maxn], mx[maxn], mi[maxn], q[maxn];
inline int read()
{
int x = 0, f = 1;
char ch = getchar();
while(ch < '0' || ch > '9')
{
if(ch == '-')
{
f = -1;
}
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = (x << 1) + (x << 3) + (ch^48);
ch = getchar();
}
return x * f;
}
void getmin()
{
int h = 1, t = 0;
for(int i=1; i<k; i++)
{
while(h <= t && a[q[t]] >= a[i]) t--;
q[++t] = i;
}
for(int i=k; i<=n; i++)
{
while(q[h] <= i - k) h++;
while(h <= t && a[q[t]] >= a[i]) t--;
q[++t] = i; mi[i] = a[q[h]];
//printf("mi[%d] = %d\n", i, mi[i]);
}
}
void getmax()
{
int h = 1, t = 0;
for(int i=1; i<k; i++)
{
while(h <= t && a[q[t]] <= a[i]) t--;
q[++t] = i;
}
for(int i=k; i<=n; i++)
{
while(q[h] <= i - k) h++;
while(h <= t && a[q[t]] <= a[i]) t--;
q[++t] = i; mx[i] = a[q[h]];
//printf("mx[%d] = %d\n", i, mx[i]);
}
}
int main()
{
n = read(); k = read();
for(int i=1; i<=n; i++) a[i] = read();
getmax(); getmin();
for(int i=k; i<=n; i++)
{
printf("%d ", mi[i]);
}
printf("\n");
for(int i=k; i<=n; i++)
{
printf("%d ", mx[i]);
}
printf("\n");
return 0;
}树状数组——区间修改，区间查询
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 3;
int n, m, a[maxn];
ll c1[maxn], c2[maxn];
char op[7];
inline ll read()
{
ll x = 0, f = 1;
char ch = getchar();
while(ch < '0' || ch > '9')
{
if(ch == '-')
{
f = -1;
}
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = (x << 1) + (x << 3) + (ch^48);
ch = getchar();
}
return x * f;
}
void add(int x, ll d)
{
int i = x;
while(i <= n)
{
c1[i] += d;
c2[i] += x * d;//注意是x*d而不是i*d
i += (i & -i);
}
}
ll query(int x)
{
ll ans = 0;
int i = x;
while(i > 0)
{
ans += (x+1)*c1[i]-c2[i];
i -= (i & -i);//x和i是不同的，从表达式里就可以看出来
}
return ans;
}
int main()
{
n = read();
for(int i=1; i<=n; i++)
{
a[i] = read();
add(i, a[i]-a[i-1]);
}
m = read();
for(int i=1; i<=m; i++)
{
scanf("%s", op);
if(op[0] == 'A')
{
int l = read(), r = read(); ll d = read();
add(l, d); add(r+1, -d);
}
else
{
int l = read(), r = read();
if(l == 1) printf("%lld\n", query(r));
else printf("%lld\n", query(r) - query(l-1));
}
}
return 0;
}二维树状数组——移动电话
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1500;
int fl, s, c[maxn][maxn];
inline int read()
{
int x = 0, f = 1;
char ch = getchar();
while(ch < '0' || ch > '9')
{
if(ch == '-')
{
f = -1;
}
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = (x << 1) + (x << 3) + (ch^48);
ch = getchar();
}
return x * f;
}
void add(int x, int y, int d)
{
for(int i=x; i<=s; i+=(i&-i))
{
for(int j=y; j<=s; j+=(j&-j))
{
c[i][j] += d;
}
}
}
int query(int x, int y)
{
int ans = 0;
for(int i=x; i>0; i-=(i&-i))
{
for(int j=y; j>0; j-=(j&-j))
{
ans += c[i][j];
}
}
return ans;
}
int main()
{
fl = read(); s = read();
fl = read();
while(fl != 3)
{
if(fl == 1)
{
int x = read()+1, y = read()+1, d = read();
add(x, y, d);
}
else if(fl == 2)
{
int l = read()+1, b = read()+1, k = read()+1, t = read()+1;
int ans = query(k, t)-query(k, b-1)-query(l-1, t)+query(l-1, b-1);
printf("%d\n", ans);
}
fl = read();
}
return 0;
}线段树——最大值
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e6 + 3;
int n, m, a[maxn];
inline ll read()
{
ll x = 0, f = 1;
char ch = getchar();
while(ch < '0' || ch > '9')
{
if(ch == '-')
{
f = -1;
}
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = (x << 1) + (x << 3) + (ch^48);
ch = getchar();
}
return x * f;
}
struct node
{
struct tree
{
int max;
}t[maxn<<2];
void pushup(int x)
{
t[x].max = max(t[x<<1].max, t[x<<1|1].max);
}
void build(int x, int l, int r)
{
if(l == r)
{
t[x].max = a[l];
return;
}
int mid = (l + r) >> 1;
build(x<<1, l, mid);
build(x<<1|1, mid+1, r);
pushup(x);
}
int query(int x, int l, int r, int L, int R)
{
if(L <= l && r <= R)
{
return t[x].max;
}
int mid = (l + r) >> 1, ans = 0;
if(L <= mid) ans = max(ans, query(x<<1, l, mid, L, R));
if(R > mid) ans = max(ans, query(x<<1|1, mid+1, r, L, R));
return ans;
}
}t;
int main()
{
n = read();
for(int i=1; i<=n+1; i++) a[i] = read();
t.build(1, 1, n+1);
m = read();
for(int i=1; i<=m; i++)
{
int l = read()+1, r = read()+1;
int ans = t.query(1, 1, n+1, l, r);
printf("%d\n", ans);
}
return 0;
}线段树——区间求和
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 3;
int n, m;
ll a[maxn];
char op[5];
inline ll read()
{
ll x = 0, f = 1;
char ch = getchar();
while(ch < '0' || ch > '9')
{
if(ch == '-')
{
f = -1;
}
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = (x << 1) + (x << 3) + (ch^48);
ch = getchar();
}
return x * f;
}
struct node
{
struct tree
{
ll sum, lazy;
}t[maxn<<2];
void pushup(int x)
{
t[x].sum = t[x<<1].sum + t[x<<1|1].sum;
}
void pushdown(int x, int l, int r)
{
int ls = x<<1, rs = x<<1|1, mid = (l + r) >> 1;
ll lz = t[x].lazy; t[x].lazy = 0;
t[ls].lazy += lz;
t[ls].sum += lz * (mid - l + 1);
t[rs].lazy += lz;
t[rs].sum += lz * (r - mid);
}
void build(int x, int l, int r)
{
if(l == r)
{
t[x].sum = a[l];
return;
}
int mid = (l + r) >> 1;
build(x<<1, l, mid);
build(x<<1|1, mid+1, r);
pushup(x);
}
void update(int x, int l, int r, int L, int R, ll d)
{
if(L <= l && r <= R)
{
t[x].sum += (r - l + 1) * d; t[x].lazy += d;
return;
}
if(t[x].lazy) pushdown(x, l, r);
int mid = (l + r) >> 1;
if(L <= mid) update(x<<1, l, mid, L, R, d);
if(R > mid) update(x<<1|1, mid+1, r, L, R, d);
pushup(x);
}
ll query(int x, int l, int r, int L, int R)
{
if(L <= l && r <= R)
{
return t[x].sum;
}
if(t[x].lazy) pushdown(x, l, r);
int mid = (l + r) >> 1; ll ans = 0;
if(L <= mid) ans += query(x<<1, l, mid, L, R);
if(R > mid) ans += query(x<<1|1, mid+1, r, L, R);
return ans;
}
}t;
int main()
{
n = read();
for(int i=1; i<=n; i++) a[i] = read();
t.build(1, 1, n);
m = read();
for(int i=1; i<=m; i++)
{
scanf("%s", op);
if(op[0] == 'A')
{
int l = read(), r = read(); ll d = read();
t.update(1, 1, n, l, r, d);
}
else
{
int l = read(), r = read();
ll ans = t.query(1, 1, n, l, r);
printf("%lld\n", ans);
}
}
return 0;
}树状数组——区间修改，单点查询
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 3;
int n, m, a[maxn];
ll c[maxn];
char op[7];
inline ll read()
{
ll x = 0, f = 1;
char ch = getchar();
while(ch < '0' || ch > '9')
{
if(ch == '-')
{
f = -1;
}
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = (x << 1) + (x << 3) + (ch^48);
ch = getchar();
}
return x * f;
}
void add(int x, ll d)
{
while(x <= n)
{
c[x] += d;
x += (x & -x);
}
}
ll query(int x)
{
ll ans = 0;
while(x > 0)
{
ans += c[x];
x -= (x & -x);
}
return ans;
}
int main()
{
n = read();
for(int i=1; i<=n; i++) a[i] = read();
add(1, a[1]);
for(int i=2; i<=n; i++) add(i, a[i]-a[i-1]);
m = read();
for(int i=1; i<=m; i++)
{
scanf("%s", op);
if(op[0] == 'A')
{
int l = read(), r = read(); ll d = read();
add(l, d); add(r+1, -d);
}
else
{
int p = read();
ll ans = query(p);
printf("%lld\n", ans);
}
}
return 0;
}HH的项链——树状数组
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e6 + 3;
const int maxm = 2e5 + 3;
int n, m, R, a[maxn], ans[maxm], las[maxn], c[maxn];
inline int read()
{
int x = 0, f = 1;
char ch = getchar();
while(ch < '0' || ch > '9')
{
if(ch == '-')
{
f = -1;
}
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = (x << 1) + (x << 3) + (ch^48);
ch = getchar();
}
return x * f;
}
struct que
{
int l, r, id;
bool operator < (const que &T) const
{
return r < T.r;
}
}q[maxm];
inline void add(int x, int d)
{
while(x <= n)
{
c[x] += d;
x += (x & -x);
}
}
inline int query(int x)
{
int ans = 0;
while(x > 0)
{
ans += c[x];
x -= (x & -x);
}
return ans;
}
int main()
{
n = read();
for(int i=1; i<=n; i++) a[i] = read();
m = read();
for(int i=1; i<=m; i++)
{
q[i].l = read(); q[i].r = read(); q[i].id = i;
}
sort(q+1, q+1+m);
for(int i=1; i<=m; i++)
{
while(R < q[i].r)
{
R++;
if(las[a[R]]) add(las[a[R]], -1);
add(R, 1);
las[a[R]] = R;
}
if(q[i].l == 1) ans[q[i].id] = query(R);
else ans[q[i].id] = query(R)-query(q[i].l-1);
}
for(int i=1; i<=m; i++)
{
printf("%d\n", ans[i]);
}
return 0;
}dfs序1
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e6 + 3;
int dfn[maxn], v[maxn], nt, n, m, r, siz[maxn], fnd[maxn];
ll c[maxn];
inline int read()
{
int x = 0, f = 1;
char ch = getchar();
while(ch < '0' || ch > '9')
{
if(ch == '-')
{
f = -1;
}
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = (x << 1) + (x << 3) + (ch^48);
ch = getchar();
}
return x * f;
}
struct node
{
int nxt, to;
}a[maxn<<1];
int head[maxn], len;
void add(int x, int y)
{
a[++len].to = y; a[len].nxt = head[x];
head[x] = len;
}
void dfs(int u, int fa)
{
dfn[++nt] = u; fnd[u] = nt;
siz[u] = 1;
for(int i=head[u]; i; i=a[i].nxt)
{
int v = a[i].to;
if(v == fa) continue;
dfs(v, u);
siz[u] += siz[v];
}
}
void addt(int x, ll d)
{
while(x <= n)
{
c[x] += d;
x += (x & -x);
}
}
ll query(int x)
{
ll ans = 0;
while(x > 0)
{
ans += c[x];
x -= (x & -x);
}
return ans;
}
int main()
{
n = read(); m = read(); r = read();
for(int i=1; i<=n; i++) v[i] = read();
for(int i=1; i<n; i++)
{
int x = read(), y = read();
add(x, y); add(y, x);
}
dfs(r, 0);
for(int i=1; i<=n; i++)
{
addt(i, v[dfn[i]]);
}
for(int i=1; i<=m; i++)
{
int fl = read();
if(fl == 1)
{
int a = read(), x = read();
addt(fnd[a], x);//不用dfn数组
}
else
{
int a = read();
ll ans = query(fnd[a]+siz[a]-1)-query(fnd[a]-1);//不用dfn数组
printf("%lld\n", ans);
}
}
return 0;
}倍增LCA——牧场旅行
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1130;
int n, q, f[maxn][12], c[maxn][12], dep[maxn];
inline int read()
{
int x = 0, f = 1;
char ch = getchar();
while(ch < '0' || ch > '9')
{
if(ch == '-')
{
f = -1;
}
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = (x << 1) + (x << 3) + (ch^48);
ch = getchar();
}
return x * f;
}
struct node
{
int nxt, to, w;
}a[maxn<<1];
int head[maxn], len;
void add(int x, int y, int w)
{
a[++len].to = y; a[len].nxt = head[x]; a[len].w = w;
head[x] = len;
}
void dfs(int x, int fa)
{
f[x][0] = fa;
dep[x] = dep[fa] + 1;
for(int i=1; i<12; i++)
{
f[x][i] = f[f[x][i-1]][i-1];
c[x][i] = c[f[x][i-1]][i-1] + c[x][i-1];
}
for(int i=head[x]; i; i=a[i].nxt)
{
int v = a[i].to;
if(v == fa) continue;
c[v][0] = a[i].w;
dfs(v, x);
}
}
int query(int x, int y)
{
if(dep[x] > dep[y]) swap(x, y);
int tmp = dep[y] - dep[x], ans = 0;
for(int i=0; tmp; i++,tmp>>=1)
{
if(tmp & 1) ans += c[y][i], y = f[y][i];
}
if(y == x) return ans;//LCA return x;
for(int i=11; i>=0&&x!=y; i--)
{
if(f[x][i] != f[y][i])
{
ans += c[x][i] + c[y][i];
x = f[x][i]; y = f[y][i];//因为写成了y = c[y][i]错了3次！？
}
}
ans += c[x][0] + c[y][0];
return ans;//LCA return f[x][0]
}
int main()
{
n = read(); q = read();
for(int i=1; i<n; i++)
{
int x = read(), y = read(), w = read();
add(x, y, w); add(y, x, w);
}
dfs(1, 0);
for(int i=1; i<=q; i++)
{
int x = read(), y = read();
printf("%d\n", query(x, y));
}
return 0;
}树链剖分——路径求和+最大值
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 3e4 + 2;
int n, m, w[maxn], fa[maxn];
int dep[maxn], dfn[maxn], nt, fnd[maxn], siz[maxn], son[maxn], top[maxn];
char op[8];
ll ans;
inline int read()
{
int x = 0, f = 1;
char ch = getchar();
while(ch < '0' || ch > '9')
{
if(ch == '-')
{
f = -1;
}
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = (x << 1) + (x << 3) + (ch^48);
ch = getchar();
}
return x * f;
}
struct edge
{
int nxt, to;
}a[maxn<<1];
int head[maxn], len;
void add(int x, int y)
{
a[++len].to = y; a[len].nxt = head[x];
head[x] = len;
}
void dfs1(int u, int fat, int depth)
{
fa[u] = fat;
dep[u] = depth;
siz[u] = 1;
son[u] = 0;
int maxsize = 0;
for(int i=head[u]; i; i=a[i].nxt)
{
int v = a[i].to;
if(v == fat) continue;
dfs1(v, u, depth+1);
siz[u] += siz[v];
if(siz[v] > maxsize)
{
maxsize = siz[v];
son[u] = v;
}
}
}
void dfs2(int u, int ancestor)
{
dfn[u] = ++nt; fnd[nt] = u;
top[u] = ancestor;
if(son[u])
{
dfs2(son[u], ancestor);
}
for(int i=head[u]; i; i=a[i].nxt)
{
int v = a[i].to;
if(v == fa[u] || v == son[u]) continue;
dfs2(v, v);
}
}
struct node
{
struct tree
{
int max; ll sum;
}t[maxn<<2];
void pushup(int x)
{
t[x].max = max(t[x<<1].max, t[x<<1|1].max);
t[x].sum = t[x<<1].sum + t[x<<1|1].sum;
}
void build(int x, int l, int r)
{
if(l == r)
{
t[x].max = t[x].sum = w[fnd[l]];
return;
}
int mid = (l + r) >> 1;
build(x<<1, l, mid);
build(x<<1|1, mid+1, r);
pushup(x);
}
void update(int x, int l, int r, int p, int v)
{
if(l == r)
{
t[x].max = t[x].sum = v;
return;
}
int mid = (l + r) >> 1;
if(p <= mid) update(x<<1, l, mid, p, v);
else update(x<<1|1, mid+1, r, p, v);
pushup(x);
}
int querymax(int x, int l, int r, int L, int R)
{
if(L <= l && r <= R)
{
return t[x].max;
}
int mid = (l + r) >> 1, ans = -1e9;
if(L <= mid) ans = querymax(x<<1, l, mid, L, R);
if(R > mid) ans = max(ans, querymax(x<<1|1, mid+1, r, L, R));
return ans;
}
ll querysum(int x, int l, int r, int L, int R)
{
if(L <= l && r <= R)
{
return t[x].sum;
}
int mid = (l + r) >> 1; ll ans = 0;
if(L <= mid) ans = querysum(x<<1, l, mid, L, R);
if(R > mid) ans += querysum(x<<1|1, mid+1, r, L, R);
return ans;
}
}t;
int querymax(int x, int y)
{
int ans = -1e9;
while(top[x] != top[y])
{
if(dep[top[x]] < dep[top[y]]) swap(x, y);
ans = max(ans, t.querymax(1, 1, n, dfn[top[x]], dfn[x]));
x = fa[top[x]];
}
if(dep[x] > dep[y]) swap(x, y);
ans = max(ans, t.querymax(1, 1, n, dfn[x], dfn[y]));
return ans;
}
ll querysum(int x, int y)
{
ll ans = 0;
while(top[x] != top[y])
{
if(dep[top[x]] < dep[top[y]]) swap(x, y);
ans += t.querysum(1, 1, n, dfn[top[x]], dfn[x]);
x = fa[top[x]];
}
if(dep[x] > dep[y]) swap(x, y);
ans += t.querysum(1, 1, n, dfn[x], dfn[y]);
return ans;
}
int main()
{
n = read();
for(int i=1; i<n; i++)
{
int x = read(), y = read();
add(x, y); add(y, x);
}
dfs1(1, 1, 1);
dfs2(1, 1);
for(int i=1; i<=n; i++) w[i] = read();
t.build(1, 1, n);
m = read();
while(m--)
{
scanf("%s", op);
if(op[0] == 'C')
{
int x = read(), d = read();
t.update(1, 1, n, dfn[x], d);
}
else if(op[1] == 'M')
{
int l = read(), r = read();
ans = querymax(l, r);
printf("%lld\n", ans);
}
else
{
int l = read(), r = read();
ans = querysum(l, r);
printf("%lld\n", ans);
}
}
return 0;
}正反图Spfa——奶牛派对
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1303;
const int M = 1e6 + 2;
int n, m, q, dis[maxn][2], ans;
bool vis[maxn];
inline int read()
{
int x = 0, f = 1;
char ch = getchar();
while(ch < '0' || ch > '9')
{
if(ch == '-')
{
f = -1;
}
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = (x << 1) + (x << 3) + (ch^48);
ch = getchar();
}
return x * f;
}
struct edge
{
int nxt, to, w;
}a[M][2];
int head[maxn][2], len[2];
void add(int x, int y, int w, int c)
{
a[++len[c]][c].to = y; a[len[c]][c].nxt = head[x][c]; a[len[c]][c].w = w;
head[x][c] = len[c];
}
queue<int> que;
void spfa(int s, int c)
{
memset(vis, 0, sizeof(vis));
que.push(s);
vis[s] = 1;
dis[s][c] = 0;
while(!que.empty())
{
int x = que.front(); que.pop();
vis[x] = 0;
for(int i=head[x][c]; i; i=a[i][c].nxt)
{
int y = a[i][c].to;
if(dis[y][c] > dis[x][c] + a[i][c].w)
{
dis[y][c] = dis[x][c] + a[i][c].w;
if(!vis[y])
{
que.push(y);
}
}
}
}
}
int main()
{
n = read(); m = read(); q = read();
for(int i=1; i<=m; i++)
{
int x = read(), y = read(), w = read();
add(x, y, w, 0); add(y, x, w, 1);
}
memset(dis, 0x3f, sizeof(dis));
spfa(q, 0); spfa(q, 1);
for(int i=1; i<=n; i++)
{
ans = max(ans, dis[i][0] + dis[i][1]);
}
printf("%d\n", ans);
return 0;
}带权并查集——银河英雄传说
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 3e4 + 4;
int T, fa[maxn], fnt[maxn], num[maxn];
char op[3];
inline int read()
{
int x = 0, f = 1;
char ch = getchar();
while(ch < '0' || ch > '9')
{
if(ch == '-')
{
f = -1;
}
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = (x << 1) + (x << 3) + (ch^48);
ch = getchar();
}
return x * f;
}
int find(int x)
{
if(x == fa[x]) return x;
//else return fa[x] = find(fa[x]);
int fx = find(fa[x]);
fnt[x] += fnt[fa[x]];//not fx，一次只跳一步
fa[x] = fx;
return fa[x];
}
int main()
{
T = read();
for(int i=1; i<=30000; i++)
{
fa[i] = i; num[i] = 1; fnt[i] = 0;
}
while(T--)
{
scanf("%s", op);
int x = read(), y = read();
int fx = find(x), fy = find(y);
if(op[0] == 'M')
{
fnt[fx] += num[fy];
num[fy] += num[fx];
num[fx] = 0;
fa[fx] = fy;
}
else
{
if(fx != fy) printf("-1\n");
else printf("%d\n", abs(fnt[x]-fnt[y])-1);
}
}
return 0;
}RMQ_ST表
void RMQ_init()
{
for(int i=1; i<=n; i++) d[i][0] = a[i];
for(int j=1; (1<<j)<=n; j++)
{
for(int i=1; i+(1<<j)-1<=n; i++)
{
d[i][j] = min(d[i][j-1], d[i+(1<<(j-1))][j-1]);
}
}
}
int RMQ(int L, int R)
{
int k = 0;
while((1<<(k+1)) <= R-L+1) k++;
return min(d[L][k], d[R-(1<<k)+1][k]);
}tarjan缩点（有向图的强连通分量）
void tarjan(int x)
{
dfn[x] = low[x] = ++num;
stk.push(x);
v[x] = 1;
vis[x] = 1;
for(int i=head[x]; i; i=a[i].nxt)
{
if(!dfn[a[i].to])
{
tarjan(a[i].to);
low[x] = min(low[x], low[a[i].to]);
}
else if(v[a[i].to])
{
low[x] = min(low[x], dfn[a[i].to]);
}
}
if(low[x] == dfn[x])
{
int y, tot = 0;
++t;
do
{
y = stk.top();
stk.pop();
v[y] = 0;
belong[y] = t;
tot++;
}while(y != x);
if(tot > 1) ans = min(ans, tot);
}
}tarjan求割点（无向图）
void tarjan(int x)
{
dfn[x] = low[x] = ++num;
int son = 0;
for(int i=head[x]; i; i=a[i].nxt)
{
if(!dfn[a[i].to])
{
son++;
tarjan(a[i].to);
low[x] = min(low[x], low[a[i].to]);
if(dfn[x] <= low[a[i].to])
{
if(x != root || son > 1) cut[x] = true;
}
}
else
{
low[x] = min(low[x], dfn[a[i].to]);
}
}
}tarjan求割边（无向图）
void tarjan(int now, int fa)
{
dfn[now] = low[now] = ++num;
bool first = true;
for(int i=head[now]; i; i=a[i].nxt)
{
if(first && a[i].to == fa)//first不是没有用，它管跳过第一次，考虑重边了
{
first = false;
continue;
}
if(!dfn[to[i]])
{
tarjan(to[i]);
low[now] = min(low[now], low[a[i].to]);
if(dfn[now] < dfn[a[i].to])
{
cnt++;
cut[i] = cut[i^1] = true;
}
}
else
{
low[now] = min(low[now], dfn[a[i].to]);
}
}
}tarjan点双连通（无向图）
void tarjan(int now, int fa)
{
vis[now] = true;
dfn[now] = low[now] = ++dfs_clock;
stk[++top] = now;
bool first = true;
int son = 0;
for(int i=head[now]; i; i=a[i].nxt)
{
int to = a[i].to;
if(first && to == fa)
{
first = false;
continue;
}
if(!dfn[to])
{
son++;
tarjan(to, now);
low[now] = min(low[now], low[to]);
if(dfn[now] <= low[to])
{
cut[now] = true;
cnt++;
dslt[cnt].push_back(now);//既然不出栈就提前放进去
int x;
do
{
x = stk[top--];
dslt[cnt].push_back(x);
}while(x != to);//now不能直接出栈，它有可能属于多个点双
}
}
else
{
low[now] = min(low[now], dfn[to]);
}
}
if(fa == -1 && son == 1) cut[now] = false;
}Tarjan边双联通（无向图）
void tarjan(int now, int eid)
{
vis[now] = true;
dfn[now] = low[now] = ++dfs_clock;
stk[++top] = now;
for(int i=head[now]; i; i=a[i].nxt)
{
int to = a[i].to;
int id = a[i].id;
if(id == eid^1) continue;
if(!dfn[to])
{
tarjan(to, id);
low[now] = min(low[now], low[to]);
}
else
{
low[now] = min(low[now], dfn[to]);
}
}
if(dfn[now] == low[now])
{
cut[eid] = true;
cnt++;
int x;
do
{
x = stk[top--];
bslt[cnt].push_back(x);
}while(x != now);
}
}二分图
bool dfs(int x)
{
for(int i=1; i<=n; i++)
{
if(!vis[i] && mp[x][i])
{
vis[i] = 1;
if(match[i] == 0 || dfs(match[i]))
{
match[i] = x;
return true;
}
}
}
return false;
}
int main()
{
n = read(); k = read();
for(int i=1; i<=k; i++)
{
int x = read(), y = read();
mp[x][y] = 1;
}
for(int i=1; i<=n; i++)
{
memset(vis, 0, sizeof(vis));
if(dfs(i)) ans++;
}
printf("%d", ans);
return 0;
}重载运算符
struct node
{
int r;
bool operator < (const node &a) const
{
return r < a.r;//return r > a.r;优先队列小的优先出队
}
}p[maxn];
//如果使用sort得到按照r从小到大排序的p
//优先队列是相反的lazy
#include<bits/stdc++.h>
using namespace std;
#define maxn 100000
#define lson (rt << 1)
#define rson (rt << 1 | 1)
long long n, m, a[maxn], u, t, v, ans;
char s[7];
struct node
{
long long l, r, sum, lazy;//结构体里的数据类型别忘了改！
};
node tree[maxn * 4];
long long read()
{
long long x = 0, f = 1;
char ch = getchar();
while(ch > '9' || ch < '0')
{
if(ch == '-')
{
f = -1;
}
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = (x << 1)+ (x << 3) + (ch^48);
ch = getchar();
}
return x * f;
}
void pushup(long long rt)
{
tree[rt].sum = tree[lson].sum + tree[rson].sum;
//tree[rt].min = min(tree[lson].min, tree[rson].min);
}
void build(long long rt, long long l, long long r)
{
tree[rt].l = l;
tree[rt].r = r;
if(l == r)
{
tree[rt].sum = a[l];
return;
}
long long mid = (l + r) >> 1;
build(lson, l, mid);
build(rson, mid+1, r);
pushup(rt);
}
void pushdown(long long rt)
{
if(tree[rt].lazy)
{
long long lz = tree[rt].lazy;
tree[rt].lazy = 0;
tree[lson].lazy += lz;
tree[rson].lazy += lz;
tree[lson].sum += lz * (tree[lson].r - tree[lson].l + 1);
tree[rson].sum += lz * (tree[rson].r - tree[rson].l + 1);
}
}
void update(long long rt, long long l, long long r, long long val)
{
if(l <= tree[rt].l && tree[rt].r <= r)
{
tree[rt].lazy += val;
tree[rt].sum += val * (tree[rt].r - tree[rt].l + 1);
return;
}
pushdown(rt);
int mid = (tree[rt].l + tree[rt].r) >> 1;
if(l <= mid) update(lson, l, r, val);
if(r > mid) update(rson, l, r, val);
pushup(rt);
}
long long query(long long rt, long long l, long long r)
{
if(l <= tree[rt].l && tree[rt].r <= r)
{
return tree[rt].sum;
}
pushdown(rt);
long long mid = (tree[rt].l + tree[rt].r) >> 1;
long long val = 0;
if(l <= mid) val += query(lson, l, r);
if(r > mid) val += query(rson, l, r);
return val;
}
/*void update(int rt, int pos, int val)
{
if(tree[rt].l == tree[rt].r)
{
tree[rt].sum += val;
tree[rt].min += val;
return;
}
int mid = (tree[rt].l + tree[rt].r) >> 1;
if(pos <= mid) update(lson, pos, val);
else update(rson, pos, val);
pushup(rt);
}*/
/*int query(int rt, int l, int r)
{
if(l <= tree[rt].l && tree[rt].r <= r)
{
return tree[rt].sum;
}
int mid = (tree[rt].l + tree[rt].r) >> 1;
if(r <= mid) return query(lson, l, r);
else if(l > mid) return query(rson, l, r);
else return (query(lson, l, r) + query(rson, l, r));
}*/
/*int monn(int rt, int l, int r)
{
if(l <= tree[rt].l && tree[rt].r <= r)
{
return tree[rt].min;
}
int mid = (tree[rt].l + tree[rt].r) >> 1;
int mi = 0x7fffffff;
if(l <= mid) mi = min(mi, monn(lson, l ,r));
if(r > mid) mi = min(mi, monn(rson, l, r));
return mi;
}*/
int main()
{
n = read();
if(n == 0) exit(0);
for(int i=1; i<=n; i++)
{
a[i] = read();
}
build(1, 1, n);
m = read();
for(int i=1; i<=m; i++)
{
scanf("%s", s);
u = read();
if(s[0] == 'A')
{
t = read();
v = read();
update(1, u, t, v);
}
if(s[0] == 'Q')
{
ans = query(1, u, u);
printf("%lld\n", ans);
}
}
}混合背包
#include<iostream>
#include<algorithm>
#include<iomanip>
#include<cmath>
#include<stack>
#include<map>
#include<queue>
#include<cstring>
#include<cstdio>
#include<vector>
#include<cstdio>
#include<deque>
#include<bitset>
using namespace std;
#define maxn 300000
int n, m;
int f[maxn], c[maxn];
int v[maxn], p[maxn];
inline int read()
{
int x = 0, f = 1;
char ch = getchar();
while(ch > '9' || ch < '0')
{
if(ch == '-')
{
f = -1;
}
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = (x << 1) + (x << 3) + (ch^48);
ch = getchar();
}
return x * f;
}
int main()
{
m = read(); n =read();
for(int i=1; i<=n; i++)
{
v[i] = read();
c[i] = read();
p[i] = read();
}
for(int i=1; i<=n; i++)
{
if(p[i] == 0)
{
for(int j=v[i]; j<=m; j++)
{
f[j] = max(f[j], f[j-v[i]]+c[i]);
}
}
else
{
for(int k=1; k<=p[i]; k<<=1)
{
for(int j=m; j>=v[i]*k; j--)
{
f[j] = max(f[j], f[j-v[i]*k]+c[i]*k);
}
p[i] -= k;
}
if(p[i])
{
for(int j=m; j>=v[i]*p[i]; j--)
{
f[j] = max(f[j], f[j-v[i]*p[i]]+c[i]*p[i]);
}
}
}
}
printf("%d", f[m]);
return 0;
}Hash
#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long ll;
const int maxn = 1e6 + 7;
const int M = 998244353;
const int B = 233;
char w[maxn], t[maxn];
int n, len1, len2, ans;
ll h1, f[maxn], p[maxn], h2;
int main()
{
p[0] = 1;
for(int i=1; i<=10000; i++)
{
p[i] = p[i-1] * B;
}
scanf("%d", &n);
while(n--)
{
ans = 0; h1 = 0;
memset(f, 0, sizeof(f));
scanf("%s%s", w, t);
len1 = strlen(w);
len2 = strlen(t);
for(int i=0; i<len1; i++)
{
h1 = h1 * B + w[i];
}
f[0] = t[0];
for(int i=1; i<len2; i++)
{
f[i] = f[i-1] * B + t[i];
}
h2 = f[len1-1];
if(h1 == h2) ans++;
for(int i=1; i+len1-1<len2; i++)
{
h2 = f[i+len1-1] - f[i-1]*p[len1];
if(h1 == h2) ans++;
}
printf("%d\n",ans);
}
return 0;
}KMP
int main()
{
//找到b在a中出现的所有位置
scanf("%s%s", a+1, b+1);
int la = strlen(a+1), lb = strlen(b+1);
int j = 0;
p[1] = 0;
for(int i=2; i<=lb; i++)
{
while(j > 0 && b[i] != b[j+1]) j = p[j];
if(b[i] == b[j+1]) j++;
p[i] = j;
}
j = 0;
for(int i=1; i<=la; i++)
{
while(j > 0 && a[i] != b[j+1]) j = p[j];
if(a[i] == b[j+1]) j++;
if(j == lb) printf("%d\n", i-lb+1), j = p[j];
}
return 0;
}Trie
int n, m, ch[maxn*50][26], tag[N], tot=1;
int main()
{
scanf("%d", &n);
for(int i=1; i<=n; i++)
{
scanf("%s", s+1);
int u = 1;
for(int j=1; s[j]; j++)
{
int c = s[j] - 'a';
if(!ch[u][c]) ch[u][c] = ++tot;
u = ch[u][c];
}
tag[u] = 1;
}
scanf("%d", &m);
while(m--)
{
scanf("%s", s+1);
int u = 1;
for(int j=1; s[j]; j++)
{
int c = s[j] - 'a';
u = ch[u][c];
if(!u) break;
}
if(tag[u] == 1)
{
tag[u] = 2;
printf("OK\n");
}
else if(tag[u] == 2)
{
printf("REPEAT\n");
}
else
{
printf("WRONG\n");
}
}
return 0;
}矩阵乘法--发展结果
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 5;
const int mod = 998244353;
ll a1, b, c, d, n;
//int n;
struct Matrix
{
//int a1, b, c, d;
ll a[N][N] = {
{1, a1, 0, 0, a1},
{0, b, 1, 0, b},
{0, c, 0, 1, c},
{0, d, 0, 0, d},
{0, 0, 0, 0, 1}
};
}tmp;
Matrix clear()
{
Matrix res;
for(int i=0; i<N; i++)
{
for(int j=0; j<N; j++)
{
res.a[i][j] = 0;
}
}
return res;
}
Matrix init()
{
Matrix res = clear();
for(int i=0; i<N; i++)
{
res.a[i][i] = 1;
}
return res;
}
Matrix operator *(Matrix x, Matrix y)
{
Matrix res = clear();
for(int i=0; i<N; i++)
{
for(int j=0; j<N; j++)
{
for(int k=0; k<N; k++)
{
res.a[i][j] = (res.a[i][j]+(ll)x.a[i][k]*y.a[k][j]%mod)%mod;
}
}
}
return res;
}
Matrix power(Matrix a, ll b)
{
Matrix res = init();
while(b)
{
if(b & 1) res = res * a;
a = a * a;
b >>= 1;
/*for(int i=0; i<N; i++)
{
for(int j=0; j<N; j++)
{
printf("%d ", res.a[i][j]);
}
printf("\n");
}*/
}
return res;
}
/*void print(Matrix t){
for(int i=0;i<N;++i)
for(int j=0;j<N;++j)
printf("%d%c",t.a[i][j]," \n"[j==N-1]);
}*/
int main()
{
scanf("%lld%lld%lld%d%lld", &a1, &b, &c, &d, &n);
tmp.a[0][1] = a1; tmp.a[0][4] = a1;
tmp.a[1][1] = b; tmp.a[1][4] = b;
tmp.a[2][1] = c; tmp.a[2][4] = c;
tmp.a[3][1] = d; tmp.a[3][4] = d;
Matrix t = power(tmp, n-1);
//f1 f1 non non f1 初始 1 1 0 0 1
ll ans = (t.a[0][4]+t.a[1][4]+t.a[4][4])%mod;
/*for(int i=0; i<N; i++)
{
for(int j=0; j<N; j++)
{
printf("%d ", t.a[i][j]);
}
printf("\n");
}*/
printf("%lld", ans);
return 0;
}类似数位DP的分类讨论
int query(int x, int lim)
{
int sum = 0, ans = 1;
int cnt[32] = {0};
for(int i=0; i<30; i++) cnt[i+1] = cnt[i] + (x>>i&1);
//在第i位更低位有几个1，不包括第i位
for(int i=30; i>=0; i--)
{
int t = x >> i & 1;
if(!t) continue;
int tmp = sum + (1 << i);
if(tmp > lim) continue;
/*
每一次加上把这一位填0的情况数，然后把这一位填上1再往下
如果这一位本来不能填1，那就留待以后再算，每次加上的都是以后不再考虑的
所以和这一位有关的以后不考虑的情况已经加上了，这一位已经填好了
就像这一位从一开始就不一样
*/
sum = tmp; ans = (ans + (1<<cnt[i])) % mod;
/*
就算后来的位不会再有1，现在填上0的这一种情况也需要记录，因为接下来这一位就填上1了
那么我填这么多1的情况怎么计算啊……
*/
}
return ans;
}卡特兰数（包括递推组合数）--绳子游戏
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 2e5 + 5;
const ll mod = 1e9 + 7;
const int INF = 0x7ffffff;
int n, typ, mub[maxn], inv[maxn], f[maxn];
inline int read()
{
int x = 0, f = 1;
char ch = getchar();
while(ch > '9' || ch < '0')
{
if(ch == '-')
{
f = -1;
}
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = (x << 1) + (x << 3) + (ch^48);
ch = getchar();
}
return x * f;
}
int qpow(int a, int b)
{
int ans = 1;
while(b)
{
if(b & 1) ans = ans*1ll*a%mod;
a = a*1ll*a%mod;
b >>= 1;
}
return ans;
}
int C(int n, int m)
{
if(n < m) return 0;
return mub[n]*1ll*inv[m]%mod*inv[n-m]%mod;
}
void init()
{
mub[0] = 1;
for(int i=1; i<maxn; i++) mub[i] = mub[i-1]*1ll*i%mod;
//for(int i=1; i<maxn; i++) printf("mub[%d] = %d\n", i, mub[i]);
inv[maxn-1] = qpow(mub[maxn-1], mod-2);
//printf("%d\n", inv[maxn-1]);
for(int i=maxn-1; i>=1; i--) inv[i-1] = inv[i]*1ll*i%mod;
}
int catalan(int n)
{
return C(n*2, n)*1ll*qpow(n+1, mod-2)%mod;
}
int main()
{
init();
n = read();
printf("%d", catalan(n));
return 0;
}AC自动机
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e6 + 6;
const int mod = 1e9 + 7;
const int INF = 2147483647;
int n, T;
char s[maxn];
inline int read()
{
int x = 0, f = 1;
char ch = getchar();
while(ch < '0' || ch > '9')
{
if(ch == '-')
{
f = -1;
}
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = (x << 1) + (x << 3) + (ch^48);
ch = getchar();
}
return x * f;
}
namespace AC
{
int tr[maxn][26], tot;
int e[maxn], fail[maxn];
void insert(char *s)
{
int u = 0;
for(int i=1; s[i]; i++)
{
if(!tr[u][s[i]-'a'])
{
tr[u][s[i]-'a'] = ++tot;
}
u = tr[u][s[i]-'a'];
}
e[u]++;
}
queue<int> q;
void build()
{
for(int i=0; i<26; i++)
{
if(tr[0][i]) q.push(tr[0][i]);
}
while(q.size())
{
int u = q.front(); q.pop();
for(int i=0; i<26; i++)
{
if(tr[u][i])
{
fail[tr[u][i]] = tr[fail[u]][i];
q.push(tr[u][i]);
}
else
{
tr[u][i] = tr[fail[u]][i];
}
}
}
}
int query(char *t)
{
int u = 0, res = 0;
for(int i=1; t[i]; i++)
{
u = tr[u][t[i]-'a'];
for(int j=u; j && e[j]!=-1; j=fail[j])
{
res += e[j], e[j] = -1;
}
}
return res;
}
void clear()
{
tot = 0;
memset(e, 0, sizeof(e));
memset(fail, 0, sizeof(fail));
}
}
int main()
{
T = read();
while(T--)
{
AC::clear();
n = read();
for(int i=1; i<=n; i++)
{
scanf("%s", s+1); AC::insert(s);
}
scanf("%s", s+1);
AC::build();
printf("%d", AC::query(s));
}
return 0;
}哈希加链表存储整数（非字符串）
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 2e6 + 3;
const int p = 13331;
inline int read()
{
int x = 0, f = 1;
char ch = getchar();
while(ch < '0' || ch > '9')
{
if(ch == '-')
{
f = -1;
}
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = (x << 1) + (x << 3) + (ch^48);
ch = getchar();
}
return x * f;
}
int e[maxn], nxt[maxn], h[p], n, len;
int main()
{
n = read();
memset(h, -1, sizeof(h));
for(int i=1; i<=n; i++)
{
int o = read(), a = read();
if(o == 1)
{
nxt[++len] = h[a%p]; h[a%p] = len; e[len] = a;
}
else if(o == 2)
{
int op = 0, j = h[a%p];
while(j != -1)
{
//printf("e[%d] = %d\n", j, e[j]);
if(e[j] == a) {op = 1; break;}
j = nxt[j];
}
if(op) printf("Yes\n");
else printf("No\n");
}
else if(o == 3)
{
int j = h[a%p];
//printf("cmp h[%d] -> %d\n", a%p, nxt[j]);
if(e[j] == a) h[a%p] = nxt[h[a%p]];
else while(j != -1)
{
if(nxt[j] != -1 && e[nxt[j]] == a)
{
nxt[j] = nxt[nxt[j]];
//printf("nxt[%d] = %d\n", j, nxt[nxt[j]]);
}
j = nxt[j];
}
}
}
return 0;
}主席树（可持久化线段树）--区间k小
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 2e5 + 3;
int n, m, a[maxn], b[maxn], rt[maxn], cnt, ans;
inline int read()
{
int x = 0, f = 1;
char ch = getchar();
while(ch < '0' || ch > '9')
{
if(ch == '-')
{
f = -1;
}
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = (x << 1) + (x << 3) + (ch^48);
ch = getchar();
}
return x * f;
}
struct seg
{
struct node
{
int sum;
}t[maxn<<5];
int lc[maxn<<5], rc[maxn<<5], tot;
void build(int &x, int l, int r)
{
x = ++tot;
if(l == r) return;
int mid = (l + r) >> 1;
build(lc[x], l, mid);
build(rc[x], mid+1, r);
}
void update(int &x, int pre, int l, int r, int pos)
{
x = ++tot; lc[x] = lc[pre]; rc[x] = rc[pre];
t[x].sum = t[pre].sum + 1;
if(l == r) return;
int mid = (l + r) >> 1;
if(pos <= mid) update(lc[x], lc[x], l, mid, pos);
else update(rc[x], rc[x], mid+1, r, pos);
}
int query(int pre, int x, int l, int r, int k)
{
if(l == r) return l;
int mid = (l + r) >> 1;
int res = t[lc[x]].sum - t[lc[pre]].sum;
if(res >= k) return query(lc[pre], lc[x], l, mid, k);
else return query(rc[pre], rc[x], mid+1, r, k-res);
}
}t;
int main()
{
n = read(); m = read();
for(int i=1; i<=n; i++)
{
a[i] = read(); b[i] = a[i];
}
sort(b+1, b+1+n);
cnt = unique(b+1, b+1+n)-b-1;
t.build(rt[0], 1, cnt);
for(int i=1; i<=n; i++)
{
int x = lower_bound(b+1, b+1+cnt, a[i])-b;
t.update(rt[i], rt[i-1], 1, cnt, x);
}
while(m--)
{
int l = read(), r = read(), k = read();
ans = t.query(rt[l-1], rt[r], 1, cnt, k);
printf("%d\n", b[ans]);
}
return 0;
}Treap
#include <bits/stdc++.h>
using namespace std;
const int SIZE = 100010;
struct Treap
{
int l, r;
int val, dat;
int cnt, size;
}a[SIZE];
int tot, root, n, INF = 0x7fffffff;
int New(int val)
{
a[++tot].val = val;
a[tot].dat = rand();
a[tot].cnt = a[tot].size = 1;
return tot;
}
void Update(int p)
{
a[p].size = a[a[p].l].size + a[a[p].r].size + a[p].cnt;
}
void Build()
{
New(-INF), New(INF);
root = 1, a[1].r = 2;
Update(root);
}
int GetRankByVal(int p, int val)
{
if(p == 0) return 1;//
if(val == a[p].val) return a[a[p].l].size + 1;
if(val < a[p].val) return GetRankByVal(a[p].l, val);
return GetRankByVal(a[p].r, val) + a[a[p].l].size + a[p].cnt;
}
int GetValByRank(int p, int rank)
{
if(p == 0) return INF;
if(a[a[p].l].size >= rank) return GetValByRank(a[p].l, rank);
if(a[a[p].l].size + a[p].cnt >= rank) return a[p].val;
return GetValByRank(a[p].r, rank - a[a[p].l].size - a[p].cnt);
}
void zig(int &p)
{
int q = a[p].l;
a[p].l = a[q].r, a[q].r = p, p = q;
Update(a[p].r), Update(p);
}
void zag(int &p)
{
int q = a[p].r;
a[p].r = a[q].l, a[q].l = p, p = q;
Update(a[p].l), Update(p);
}
void Insert(int &p, int val)
{
if(p == 0)
{
p = New(val);
return;
}
if(val == a[p].val)
{
a[p].cnt++, Update(p);
return;
}
if(val < a[p].val)
{
Insert(a[p].l, val);
if(a[p].dat < a[a[p].l].dat) zig(p);
}
else
{
Insert(a[p].r, val);
if(a[p].dat < a[a[p].r].dat) zag(p);
}
Update(p);
}
int GetPre(int val)
{
int ans = 1;
int p = root;
while(p)
{
if(val == a[p].val)
{
if(a[p].l > 0)
{
p = a[p].l;
while(a[p].r > 0) p = a[p].r;
ans = p;
}
break;
}
if(a[p].val < val && a[p].val > a[ans].val) ans = p;
p = val < a[p].val ? a[p].l : a[p].r;
}
return a[ans].val;
}
int GetNext(int val)
{
int ans = 2;
int p = root;
while(p)
{
if(val == a[p].val)
{
if(a[p].r > 0)
{
p = a[p].r;
while(a[p].l > 0) p = a[p].l;
ans = p;
}
break;
}
if(a[p].val > val && a[p].val < a[ans].val) ans = p;
p = val < a[p].val ? a[p].l : a[p].r;
}
return a[ans].val;
}
void Remove(int &p, int val)
{
if(p == 0) return;
if(val == a[p].val)
{
if(a[p].cnt > 1)
{
a[p].cnt--, Update(p);
return;
}
if(a[p].l || a[p].r)
{
if(a[p].r == 0 || a[a[p].l].dat > a[a[p].r].dat)
{
zig(p), Remove(a[p].r, val);
}
else
{
zag(p), Remove(a[p].l, val);
}
Update(p);
}
else p = 0;
return;
}
val < a[p].val ? Remove(a[p].l, val) : Remove(a[p].r, val);
Update(p);
}
int main()
{
//freopen("1.txt", "r", stdin);
Build();
scanf("%d", &n);
while(n--)
{
int opt, x;
scanf("%d%d", &opt, &x);
if(opt == 1)
{
Insert(root, x);
}
else if(opt == 2)
{
Remove(root, x);
}
else if(opt == 3)
{
//printf("last = %d\n", GetPre(x));
printf("%d\n", GetRankByVal(root, x)-1);
}
else if(opt == 4)
{
printf("%d\n", GetValByRank(root, x+1));
}
else if(opt == 5)
{
printf("%d\n", GetPre(x));
}
else if(opt == 6)
{
printf("%d\n", GetNext(x));
}
}
return 0;
}最近开始复习了，或许以后还会继续用这个平台记录一些题。曾经卷来卷去死记硬背怎么都记不住的那些板子，不知道为什么这一遍忽然理解了，当我终于有了一篇记录板子的blog时已经不再需要背，当我终于看清楚了一点点OI的面目，我已经离开这篇乐土很久了……
感觉大学生活总体上是比高中幸福的，但是在hzoi的回忆，却一次又一次让我幻想重来，就好像如果这样，我当年就不会走得那么早，就好像如果我的时间永远定格在那些日子里的任意一天，都值得让我放弃之后所有的可能性，嗯，我太恋旧了，这是弱点。
在整个大学大一范围内找不到一个能和我一起讨论编程的女生，不在高处，却不胜寂寞，好在找到陪我玩第5人格的搭子还是很容易的。
关于上一篇博客使用了某些专有名词和专有图片，我狡辩，我不玩原神（虽然有账号），因为不知道怎么在这个游戏里活下去，不是被反派制裁就是在被反派制裁的路上，我太菜了。
时光花火，水月星辰
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