Solutions - NOISG 2017 重现赛
T1
主观难度：【0】
懒的写。
T2
主观难度：【0】
其实树上倍增和树剖都能做。
这里我懒得思考了于是花了 20 min 打了一个树剖。
T3
主观难度：【1】
简单组合数学题。草因为没看见最短路径数不超过 \(2^{15}\) 卡题了。
对于每个公民算出每个点到 \(s\) 和 \(t\) 的路径数，乘起来就得到了经过一个点的总路径数，然后加起来算一下即可。
#include <bits/stdc++.h>
#define llong long long
#define N 5003
#define M 40004
using namespace std;
#define bs (1<<20)
char buf[bs], *p1, *p2;
#define gc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,bs,stdin),p1==p2)?EOF:*p1++)
template<typename T>
inline void read(T& x){
x = 0; int w = 1;
char ch = gc();
while(ch < '0' || ch > '9'){
if(ch == '-') w = -w;
ch = gc();
}
while(ch >= '0' && ch <= '9')
x = (x<<3)+(x<<1)+(ch^48), ch = gc();
x *= w;
}
template<typename T, typename ...Args>
inline void read(T& x, Args& ...y){
return read(x), read(y...);
}
int n, m, k, ans;
int to[M<<1], nxt[M<<1], head[N], gsiz;
#define mkarc(u,v) (++gsiz, to[gsiz]=v, nxt[gsiz]=head[u], head[u]=gsiz)
int dis1[N], dis2[N], cnt1[N], cnt2[N];
double e[N];
int que[N], he, ta;
inline void prework1(int s, int *dis){
for(int i = 1; i <= n; ++i) dis[i] = 1e9+7;
que[he = ta = 1] = s, dis[s] = 0;
while(he <= ta){
int u = que[he++];
for(int i = head[u]; i; i = nxt[i]){
int v = to[i];
if(dis[v] > dis[u]+1){
dis[v] = dis[u]+1;
que[++ta] = v;
}
}
}
return;
}
int vis[N];
inline void prework2(int s, int t, int *d1, int *d2, int *cnt){
for(int i = 1; i <= n; ++i) cnt[i] = vis[i] = 0;
que[he = ta = 1] = s, cnt[s] = 1;
while(he <= ta){
int u = que[he++];
for(int i = head[u]; i; i = nxt[i]){
int v = to[i];
if(d1[u]+1+d2[v] != d1[t]) continue;
cnt[v] += cnt[u];
if(!vis[v]) vis[que[++ta] = v] = true;
}
}
return;
}
inline void calc(int s, int t){
prework1(s, dis1), prework1(t, dis2);
prework2(s, t, dis1, dis2, cnt1);
prework2(t, s, dis2, dis1, cnt2);
for(int i = 1; i <= n; ++i)
e[i] += 1.0*cnt1[i]*cnt2[i]/cnt1[t];
return;
}
int main(){
freopen("in.txt", "r", stdin);
read(n, m);
for(int i = 1; i <= m; ++i){
int u, v;
read(u, v), ++u, ++v;
mkarc(u, v), mkarc(v, u);
}
read(k);
for(int i = 1; i <= k; ++i){
int s, t;
read(s, t), ++s, ++t;
calc(s, t);
}
for(int i = 1; i <= n; ++i)
if(e[ans] < e[i]) ans = i;
printf("%d", ans-1);
return 0;
}
T4
主观难度：【0】
懒的写。
#include <bits/stdc++.h>
#define llong long long
#define N 100005
using namespace std;
#define bs (1<<20)
char buf[bs], *p1, *p2;
#define gc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,bs,stdin),p1==p2)?EOF:*p1++)
template<typename T>
inline void read(T& x){
x = 0; int w = 1;
char ch = gc();
while(ch < '0' || ch > '9'){
if(ch == '-') w = -w;
ch = gc();
}
while(ch >= '0' && ch <= '9')
x = (x<<3)+(x<<1)+(ch^48), ch = gc();
x *= w;
}
template<typename T, typename ...Args>
inline void read(T& x, Args& ...y){
return read(x), read(y...);
}
int n, q;
int b[N];
struct Node{
int l, r, k;
} a[N];
int l[N], r[N];
int val[N<<2], tag[N<<2];
#define ls(x) (x<<1)
#define rs(x) (x<<1|1)
#define mid ((l+r)>>1)
inline void pushup(int x){
val[x] = min(val[ls(x)], val[rs(x)]);
return;
}
inline void addtag(int x, int k){
val[x] = max(val[x], k), tag[x] = max(tag[x], k);
return;
}
inline void pushdown(int x){
if(tag[x] == 0) return;
addtag(ls(x), tag[x]), addtag(rs(x), tag[x]);
tag[x] = 0;
return;
}
inline void modify(int L, int R, int k, int x = 1, int l = 1, int r = n){
if(L > R) return;
if(L <= l && R >= r) return addtag(x, k);
pushdown(x);
if(L <= mid) modify(L, R, k, ls(x), l, mid );
if(R > mid) modify(L, R, k, rs(x), mid+1, r);
pushup(x);
return;
}
inline int getlower(int L, int R, int k, int x = 1, int l = 1, int r = n){
if(val[x] > k) return -1;
if(l == r) return l;
pushdown(x);
if(R <= mid) return getlower(L, R, k, ls(x), l, mid );
if(L > mid) return getlower(L, R, k, rs(x), mid+1, r);
int res = getlower(L, R, k, ls(x), l, mid);
if(res > -1) return res;
else return getlower(L, R, k, rs(x), mid+1, r);
}
int main(){
// freopen("in.txt", "r", stdin);
read(n, q);
for(int i = 1; i <= q; ++i){
read(a[i].l, a[i].r, a[i].k);
++a[i].l, ++a[i].r, ++a[i].k;
modify(a[i].l, a[i].r, a[i].k);
}
sort(a+1, a+q+1, [&](Node o1, Node o2){return o1.k<o2.k;});
for(int i = 1; i <= n; ++i) l[i] = 1, r[i] = n;
for(int i = 1; i <= q; ++i){
l[a[i].k] = max(l[a[i].k], a[i].l);
r[a[i].k] = min(r[a[i].k], a[i].r);
}
for(int i = 1; i <= n; ++i){
if(l[i] > r[i]){
for(int j = 1; j <= n; ++j) printf("-1 ");
return 0;
}
int pos = getlower(l[i], r[i], i);
if(pos == -1){
for(int j = 1; j <= n; ++j) printf("-1 ");
return 0;
}
modify(pos, pos, 1e9+7);
b[pos] = i-1;
}
for(int i = 1; i <= n; ++i) printf("%d ", b[i]);
return 0;
}
T5
主观难度：【2】
何意味。
首先建重构树，然后题目就变成了单点加区间数颜色。听说可以用带修莫队搞出来 \(\mathrm O(nq^{\frac{2}{3}})\) 但是我懒。而且这么做好像不是很优的样子。
于是我们使用一种方法把这个东西变成离线动态二维数点。具体地，我们对每个位置建 \((x_i, x_i, 1)\)，且对两个颜色相同距离极小的位置建 \((x_j, x_i, -1)\)。这样的意义是取颜色相同的点里位置最大的来计数。
然后随便整一个离线动态二维数点即可，树套树或者 cdq 反正都是 \(\mathrm O(n \log^2 n)\)，没什么区别。
#include <bits/stdc++.h>
#define llong long long
#define N 100005
using namespace std;
#define bs (1<<20)
char buf[bs], *p1, *p2;
#define gc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,bs,stdin),p1==p2)?EOF:*p1++)
template<typename T>
inline void read(T& x){
x = 0; int w = 1;
char ch = gc();
while(ch < '0' || ch > '9'){
if(ch == '-') w = -w;
ch = gc();
}
while(ch >= '0' && ch <= '9')
x = (x<<3)+(x<<1)+(ch^48), ch = gc();
x *= w;
}
template<typename T, typename ...Args>
inline void read(T& x, Args& ...y){
return read(x), read(y...);
}
int n, m, q;
#define pos(x,y) ((x-1)*m+y)
int a[N], b[N], fa[N][21];
struct Node{
int x, y, val;
} tmp[N];
int f[N];
inline int find(int x){
if(x == f[x]) return x;
return f[x] = find(f[x]);
}
vector<int> G[N];
int lpos[N], rpos[N], dfcnt;
inline void dfs(int u){
lpos[u] = ++dfcnt;
for(int i = 1; i <= 20; ++i) fa[u][i] = fa[fa[u][i-1]][i-1];
for(int v : G[u]) dfs(v);
rpos[u] = dfcnt;
}
int root[N];
int val[N<<8], ls[N<<8], rs[N<<8], tsiz;
#define ls(x) (x<<1)
#define rs(x) (x<<1|1)
#define mid ((l+r)>>1)
inline int newnode(){
return ++tsiz;
}
inline void pushup(int x){
val[x] = val[ls[x]]+val[rs[x]];
return;
}
inline void addtag(int x, int k){
val[x] += k;
return;
}
inline void modify(int pos, int k, int& x, int l = 1, int r = n*m){
if(!x) x = newnode();
if(l == r) return addtag(x, k);
if(pos <= mid) modify(pos, k, ls[x], l, mid );
else modify(pos, k, rs[x], mid+1, r);
pushup(x);
return;
}
inline int query(int L, int R, int &x, int l = 1, int r = n*m){
if(!x) return 0;
if(L <= l && R >= r) return val[x];
int res = 0;
if(L <= mid) res += query(L, R, ls[x], l, mid );
if(R > mid) res += query(L, R, rs[x], mid+1, r);
return res;
}
#define lowbit(x) (x&-x)
inline void modifynode(int x, int y, int k){
for(int i = x; i <= n*m; i += lowbit(i)) modify(y, k, root[i]);
return;
}
inline int querymatrix(int x1, int y1, int x2, int y2){
int res = 0;
for(int i = x2; i; i ^= lowbit(i)) res += query(y1, y2, root[i]);
for(int i = x1-1; i; i ^= lowbit(i)) res -= query(y1, y2, root[i]);
return res;
}
set<int> col[N];
inline void modifycol(int pos, int k){
int dfn = lpos[pos];
// Remove from color
set<int>::iterator it1, it2;
it1 = it2 = col[b[pos]].lower_bound(dfn);
int pos1 = 0, pos2 = 0;
if(it1 != col[b[pos]].begin()) pos1 = *--it1;
if(++it2 != col[b[pos]].end()) pos2 = *it2;
if(pos1) modifynode(dfn, pos1, 1);
if(pos2) modifynode(pos2, dfn, 1);
if(pos1 && pos2) modifynode(pos2, pos1, -1);
col[b[pos]].erase(dfn);
b[pos] = k;
// Add to color
it1 = it2 = col[b[pos]].lower_bound(dfn);
pos1 = pos2 = 0;
if(it1 != col[b[pos]].begin()) pos1 = *--it1;
if(it2 != col[b[pos]].end()) pos2 = *it2;
if(pos1) modifynode(dfn, pos1, -1);
if(pos2) modifynode(pos2, dfn, -1);
if(pos1 && pos2) modifynode(pos2, pos1, 1);
col[b[pos]].insert(dfn);
return;
}
inline int queryint(int l, int r){
return querymatrix(l, l, r, r);
}
int main(){
// freopen("in.txt", "r", stdin);
read(n, m, q);
for(int i = 1; i <= n; ++i)
for(int j = 1; j <= m; ++j) read(a[pos(i, j)]);
for(int i = 1; i <= n; ++i)
for(int j = 1; j <= m; ++j) read(b[pos(i, j)]);
for(int i = 1; i <= n; ++i)
for(int j = 1; j <= m; ++j)
tmp[pos(i, j)] = (Node){i, j, a[pos(i, j)]};
sort(tmp+1, tmp+n*m+1, [&](Node o1, Node o2){return o1.val<o2.val;});
for(int i = 1; i <= n*m*2; ++i) f[i] = i;
for(int i = 1; i <= n*m; ++i){
int x = tmp[i].x, y = tmp[i].y;
if(x != 1 && a[find(pos(x-1,y))] < tmp[i].val) f[find(pos(x-1,y))] = fa[find(pos(x-1,y))][0] = pos(x, y);
if(y != 1 && a[find(pos(x,y-1))] < tmp[i].val) f[find(pos(x,y-1))] = fa[find(pos(x,y-1))][0] = pos(x, y);
if(x != n && a[find(pos(x+1,y))] < tmp[i].val) f[find(pos(x+1,y))] = fa[find(pos(x+1,y))][0] = pos(x, y);
if(y != m && a[find(pos(x,y+1))] < tmp[i].val) f[find(pos(x,y+1))] = fa[find(pos(x,y+1))][0] = pos(x, y);
}
int rt;
for(int i = 1; i <= n*m; ++i){
if(fa[i][0]) G[fa[i][0]].push_back(i);
else rt = i;
}
dfs(rt);
for(int i = 1; i <= n*m; ++i){
set<int>::iterator it1, it2;
it1 = it2 = col[b[i]].lower_bound(lpos[i]);
int pos1 = 0, pos2 = 0;
if(it1 != col[b[i]].begin()) pos1 = *--it1;
if(it2 != col[b[i]].end()) pos2 = *it2;
modifynode(lpos[i], lpos[i], 1);
if(pos1) modifynode(lpos[i], pos1, -1);
if(pos2) modifynode(pos2, lpos[i], -1);
if(pos1 && pos2) modifynode(pos2, pos1, 1);
col[b[i]].insert(lpos[i]);
}
a[0] = 1e9+7;
while(q--){
int op, x, y, k;
read(op, y, x, k);
if(op == 1){
modifycol(pos(x,y), k);
}
if(op == 2){
int now = pos(x, y);
if(k < a[now]){
puts("0");
continue;
}
for(int i = 20; ~i; --i)
if(a[fa[now][i]] <= k) now = fa[now][i];
printf("%d\n", queryint(lpos[now], rpos[now]));
}
}
return 0;
}
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