板子
最短路
dijkstra求最短路
//有向图
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define PII pair<ll,ll>
const int Vertex_MAXN = 1e4 + 10;
const int Edge_MAXN = 5e5 + 10;
const ll inf = 9e18;
struct node{
int v,nx;
ll w;
}edge[Edge_MAXN];
int head[Vertex_MAXN],tot,n,m,s;
ll dis[Vertex_MAXN];
inline void add(int u,int v,ll w)
{
edge[++tot].v = v;
edge[tot].w = w;
edge[tot].nx = head[u];
head[u] = tot;
}
void dij(int s) {
priority_queue<PII,vector<PII>,greater<PII>> q;
fill(dis,dis+n+1,inf);
dis[s] = 0;
q.push(PII(0,s));
while (!q.empty())
{
ll d = q.top().first;
int u = q.top().second;
q.pop();
if (d != dis[u]) continue;
for (int i = head[u];i;i = edge[i].nx)
{
PII y(d+edge[i].w,edge[i].v);
if (dis[y.second]>y.first)
{
dis[y.second] = y.first;
q.push(y);
}
}
}
}
int main()
{
scanf("%d%d%d",&n,&m,&s);
for (int i = 1,u,v,w;i<=m;i++)
{
scanf("%d%d%d",&u,&v,&w);
add(u,v,w);
}
dij(s);
for (int i = 1;i<=n;i++) printf("%lld ",dis[i]);
return 0;
}
spfa求最短路
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e4 + 10;
struct node{
int v,nx,w;
}edge[500010];
int head[MAXN],tot;
inline void add(int u,int v,int w)
{
edge[++tot].v = v;
edge[tot].nx = head[u];
edge[tot].w = w;
head[u] = tot;
}
int dis[MAXN],vis[MAXN],s;
void spfa()
{
queue<int> q;
fill(dis,dis+MAXN,2147483647);
q.push(s);
vis[s] = 1;
dis[s] = 0;
while (!q.empty())
{
int u = q.front();
q.pop();
vis[u] = 0;
for (int i = head[u];i;i = edge[i].nx)
{
int v = edge[i].v;
if (dis[v] > dis[u] + edge[i].w)
{
dis[v] = dis[u] + edge[i].w;
if (!vis[v]) vis[v] = 1,q.push(v);
}
}
}
}
int main()
{
int n,m;
scanf("%d%d%d",&n,&m,&s);
for (int i = 1,u,v,w;i<=m;i++)
{
scanf("%d%d%d",&u,&v,&w);
add(u,v,w);
}
spfa();
for (int i = 1;i<=n;i++) printf("%d ",dis[i]);
return 0;
}
网络流
dinic
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 2e3;
const int inf = 2147483647;
int s,t;
struct node{
int v,w,nx;
}edge[MAXN];
int tot,head[MAXN];
inline void add(int u,int v,int w)
{
edge[tot].v = v;
edge[tot].nx = head[u];
edge[tot].w = w;
head[u] = tot++;
}
int dis[MAXN];
queue<int> q;
bool bfs()
{
memset(dis,-1,sizeof(dis));
dis[s] = 0;
q.push(s);
while (!q.empty())
{
int u = q.front();
q.pop();
for (int i = head[u];i != -1;i = edge[i].nx)
{
int v = edge[i].v;
if (dis[v] == -1 && edge[i].w > 0)
{
dis[v] = dis[u] + 1;
q.push(v);
}
}
}
return dis[t] != -1;
}
int dfs(int u,int exp)
{
if (u == t) return exp;
int flow = 0,tmp = 0;
for (int i = head[u];i != -1;i = edge[i].nx)
{
int v = edge[i].v;
if (dis[v] == dis[u] + 1 && edge[i].w > 0)
{
tmp = dfs(v,min(exp,edge[i].w));
if (!tmp) continue;
exp -= tmp;
flow += tmp;
edge[i].w -= tmp;
edge[i^1].w += tmp;
if (!exp) break;
}
}
return flow;
}
void dinic()
{
int ans = 0;
while (bfs()) ans += dfs(s,inf);
printf("%d",ans);
}
int main()
{
int n;
scanf("%d",&n);
memset(head,-1,sizeof(head));
for (int i = 1,u,v,w;i<=n;i++)
{
scanf("%d%d%d",&u,&v,&w);
add(u,v,w);
add(v,u,0);
}
s = 65,t = 90;
dinic();
return 0;
}
二分图
二分图最大匹配
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e3;
int vis[MAXN][MAXN],cx[MAXN],cy[MAXN],n,m;
bool check[MAXN];
bool dfs(int u)
{
for (int v = 1;v<=n;v++)
{
if (vis[u][v] && !check[v])//邻接矩阵存储,如果u-v之间有一条路 而且 v没有被访问过
{
check[v] = 1;//标记v已经访问
if (cy[v] == -1 || dfs(cy[v]))
{
cx[u] = v;
cy[v] = u;
return 1;
}
}
}
return 0;
}
int maxmatch()
{
int ans = 0;
memset(cx,-1,sizeof(cx));
memset(cy,-1,sizeof(cy));
for (int i = 1;i<=n;i++)//字典序从大到小
{
if (cx[i] == -1)
{
memset(check,0,sizeof(check));
ans += dfs(i);
}
}
return ans;
}
int main()
{
cin>>n;
for (int i = 1;i<=n;i++)
for (int j = 1;j<=n;j++)
cin>>vis[i][j];
int ans = maxmatch();
cout<<ans;
}
二分图KM板子
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 3e2 + 10;
int a[MAXN][MAXN],d,n;
int c_girl[MAXN],c_boy[MAXN];//记录匹配对象
int ex_girl[MAXN],ex_boy[MAXN];//记录男生和女生的期望
bool vis_girl[MAXN],vis_boy[MAXN];//记录每一轮匹配匹配过的女生和男生
int slack[MAXN]; // 记录每个汉子如果能被妹子倾心最少还需要多少期望值
bool dfs(int u)//匈牙利算法找增广路径
{
vis_girl[u] = 1;
for (int i = 1;i<=n;i++)
{
if (vis_boy[i] == 0)//每一轮匹配 每个男生只尝试一次
{
int num = ex_girl[u] + ex_boy[i] - a[u][i];
if (num == 0)//如果符合要求
{
vis_boy[i] = 1;
if (c_boy[i] == 0 || dfs(c_boy[i])) //找到一个没有匹配的男生 或者该男生的妹子可以找到其他人
{
c_girl[u] = i;
c_boy[i] = u;
return 1;
}
}
else slack[i] = min(slack[i],num);//slack可以理解为该男生要得到女生的倾心 还需多少期望值 取最小值
}
}
return 0;
}
int KM()
{
memset(ex_girl,0,sizeof(ex_girl));// 每个女生的初始期望值是与她相连的男生最大的好感度
memset(ex_boy,0,sizeof(ex_boy));// 初始每个男生的期望值为0
memset(c_girl,0,sizeof(c_girl));// 初始每个男生都没有匹配的女生
memset(c_boy,0,sizeof(c_boy));// 初始每个女生都没有匹配的男生
for (int i = 1;i<=n;i++)
for (int j = 1;j<=n;j++)
ex_girl[i] = max(ex_girl[i],a[i][j]);// 每个女生的初始期望值是与她相连的男生最大的好感度
for (int i = 1;i<=n;i++) // 尝试为每一个女生解决归宿问题
{
fill(slack,slack + MAXN,2147483647);// 因为要取最小值 初始化为无穷大
while (true) // 为每个女生解决归宿问题的方法是 ：如果找不到就降低期望值，直到找到为止
{
memset(vis_girl,0,sizeof(vis_girl));
memset(vis_boy,0,sizeof(vis_boy));// 记录每轮匹配中男生女生是否被尝试匹配过
if (dfs(i)) break;// 找到归宿 退出
// 如果不能找到 就降低期望值
// 最小可降低的期望值
d = 2147483647;
for (int j = 1;j<=n;j++)
{
if (!vis_boy[j]) d = min(d,slack[j]);
}
for (int j = 1;j<=n;j++)
{
if (vis_girl[j] == 1) ex_girl[j] -= d;//所有访问过的女生降低期望值
if (vis_boy[j] == 1) ex_boy[j] += d;//所有访问过的男生增加期望值
else slack[j] -= d; //没有访问过的boy 因为girl们的期望值降低，距离得到女生倾心又进了一步！
}
}
}
// 匹配完成 求出所有配对的好感度的和
int res = 0;
for (int i = 1;i<=n;i++) res += a[i][c_girl[i]];
return res;
}
int main()
{
cin>>n;
for (int i = 1;i<=n;i++)
for (int j = 1;j<=n;j++)
scanf("%d",&a[i][j]);
cout<<KM()<<'\n';
}
点分治
#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int MAXN = 2e4 + 10;
struct node{
int v,w,nx;
}edge[MAXN<<1];
int tot,head[MAXN];
inline void add(int u,int v,int w)
{
edge[++tot].v = v;
edge[tot].w = w;
edge[tot].nx = head[u];
head[u] = tot;
}
int sz[MAXN],maxsubtree,root,size1,vis[MAXN];//vis标记当前这个点有没有被作为分治点
void getroot(int x,int fa)//找重心，把它作为当前树的根，是根据定义来求的
{
int maxn = 0;
sz[x] = 1;
for (int i = head[x];i;i = edge[i].nx)
{
int v = edge[i].v;
if (v == fa || vis[v]) continue;
getroot(v,x);
sz[x] += sz[v];
maxn = max(maxn,sz[v]);//记录以x为根的最大子树大小
}
maxn = max(maxn,size1 - sz[x]);//当以x为根时，其祖先结点也变成了x的子树
if (maxn < maxsubtree)//寻找最大子树最小
{
maxsubtree = maxn;
root = x;
}
}
ll ans = 0,sum[4];
void dfs1(int x,int fa,int st)//统计子树中到分治点（重心）对3取模的路径有几条
{
for (int i = head[x];i;i = edge[i].nx)
{
int v = edge[i].v;
if (v == fa || vis[v]) continue;
sum[(st + edge[i].w) % 3]++;
dfs1(v,x,(st + edge[i].w) % 3);
}
}
ll cal(int x,int st)//计算路径数量
{
sum[0] = sum[1] = sum[2] = 0;
sum[st]++;
ll res = 0;
dfs1(x,0,st);//以重心为起始点，跑其子树的dfs，得到 到重心的路径权值%3为0,1,2的 数量
res = sum[1] * sum[2] * 2 + sum[0] * sum[0];//统计答案
return res;
}
void dfs(int x)
{
ans += cal(x,0);//计算经过当前点的路径的数量
vis[x] = 1;
for (int i = head[x];i;i = edge[i].nx)
{
int v = edge[i].v;
if (vis[v]) continue;
ans -= cal(v,edge[i].w);//容斥，去掉统计答案时子树中互相组成路径没有经过重心（分治点）的路径数
//之所以要加上edge[i].w是因为 以重心为根，计算子树到重心的路径权值的时候，加上了这条边
//所以以其儿子点为根计算时也要加上这条边
maxsubtree = 2147483647; size1 = sz[v];
getroot(v,0);//寻找子树的重心
dfs(root);//以子树的重心为根分治子树
}
}
int main()
{
int n;
scanf("%d",&n);
for (int i = 1,u,v,w;i<n;i++)
{
scanf("%d%d%d",&u,&v,&w);
add(u,v,w % 3);
add(v,u,w % 3);
}
maxsubtree = 2147483647; size1 = n;
getroot(1,0);//先以1为根，寻找树的重心
dfs(root);//依照重心分治
ll t = (ll)n * (ll)n;
ll g = __gcd(ans,t);
printf("%lld/%lld",ans / g,t / g);
return 0;
}
树
树链剖分
重链剖分
#include<bits/stdc++.h>
using namespace std;
#define mem(a) memset(a,0,sizeof(a))
const int MAXN = 40100;
vector<int> g[MAXN];
int n,size1[MAXN],dfn[MAXN],top[MAXN],son[MAXN],fa[MAXN],d[MAXN],num = 0,aa[MAXN];
map<pair<int,int>,int> w;
//树链剖分
void dfs1(int x)
{
d[x] = d[fa[x]] + 1;
size1[x] = 1;
for (int i = 0;i<g[x].size();i++)
{
int v = g[x][i];
if (v == fa[x]) continue;
fa[v] = x;
dfs1(v);
size1[x] += size1[v];
if (size1[v] > size1[son[x]]) son[x] = v;
}
}
void dfs2(int x,int tp)
{
dfn[x] = ++num;
top[x] = tp;
aa[num] = w[make_pair(x,fa[x])];
if (son[x]) dfs2(son[x],tp);
for (int i = 0;i<g[x].size();i++)
if (g[x][i] != fa[x] && g[x][i] != son[x]) dfs2(g[x][i],g[x][i]);
}
struct node{
int l,r,lazy,num,nl,nr;
int mid(){ return (r - l) / 2 + l;}
void merge(node a,node b)//合并两个区间
{
num = a.num + b.num;
if (a.nr == b.nl && a.nr != -1 && b.nl != -1) num--;
nl = a.nl; nr = b.nr;
}
void overturn() {swap(nr,nl);}//将区间顺序倒置
// void write() //检验时用
// {
// printf("%d %d %d %d %d %d\n",l,r,lazy,num,nl,nr);
// }
};
struct seg_tree{
node tree[MAXN * 4];
void build(int st,int ed,int x)
//建树，lazy表示当前结点下的子节点需要更改的值，nl表示左端点的数字，nr表示右端点的数字，num表示有多少段颜色相同的
{
tree[x].l = st;
tree[x].r = ed;
tree[x].lazy = -1;
if (st == ed)
{
tree[x].nl = tree[x].nr = aa[st];
tree[x].num = 1;
if (st == 1)//建树的时候，一条边上两点，深度较大的那个点的dfn作为这条线段的dfn，所以根节点1没有对应的边
{
tree[x].nl = tree[x].nr = -1;
tree[x].num = 0;
}
return;
}
int mid = tree[x].mid();
build(st,mid,x * 2);
build(mid + 1,ed,x * 2 + 1);
tree[x].merge(tree[x * 2],tree[x * 2 + 1]);
}
void modify(int st,int ed,int x,int c)//区间修改值
{
if (tree[x].l >= st && tree[x].r <= ed)
{
tree[x].nl = tree[x].nr = c;
tree[x].num = 1;
tree[x].lazy = c;
return;
}
if (tree[x].lazy != -1)
{
tree[x*2].lazy=tree[x*2+1].lazy=tree[x*2].nl=tree[x*2].nr=tree[x*2+1].nl=tree[x*2+1].nr=tree[x].lazy;
tree[x * 2].num = tree[x * 2 + 1].num = 1;
tree[x].lazy = -1;
}
int mid = tree[x].mid();
if (mid >= st) modify(st,ed,x * 2,c);
if (ed > mid) modify(st,ed,x * 2 + 1,c);
tree[x].merge(tree[x * 2],tree[x * 2 + 1]);
}
node query(int st,int ed,int x,char c)//查询区间
{
if (tree[x].l >= st && tree[x].r <= ed)
{
node temp = tree[x];
if (c == 'L') temp.overturn();//如果是从起始点开始合并
return temp;
}
if (tree[x].lazy != -1)
{
tree[x*2].lazy=tree[x*2+1].lazy=tree[x*2].nl=tree[x*2].nr=tree[x*2+1].nl=tree[x*2+1].nr=tree[x].lazy;
tree[x * 2].num = tree[x * 2 + 1].num = 1;
tree[x].lazy = -1;
}
int mid = tree[x].mid();
node ans;
ans.nl = -1; ans.nr = -1; ans.num = 0;
if (mid >= st) ans = query(st,ed,x * 2,c);
if (ed > mid)
{
if (ans.num == 0) ans = query(st,ed,x * 2 + 1,c);
else if (c == 'L') ans.merge(query(st,ed,x * 2 + 1,c),ans);
else if (c == 'R') ans.merge(ans,query(st,ed,x * 2 + 1,c));
}
tree[x].merge(tree[x * 2],tree[x * 2 + 1]);
return ans;
}
}seg;
int mapping(int x,int y,int c)//c表示当前是修改还是查询
{
int fx = top[x],fy = top[y];
node ans1,ans2;
ans1.nl = ans1.nr = ans2.nl = ans2.nr = -1;//ans1表示从起始点开始合并，ans2表示从终点开始合并
ans1.num = ans2.num = 0;
while (fx != fy)
{
if (d[fx] > d[fy])
{
if (c != -1) seg.modify(dfn[fx],dfn[x],1,c);
else
{
if (ans1.num == 0) ans1 = seg.query(dfn[fx],dfn[x],1,'L'); //如果是从起点开始合并
else ans1.merge(ans1,seg.query(dfn[fx],dfn[x],1,'L'));
}
x = fa[fx];
fx = top[x];
}
else
{
if (c != -1) seg.modify(dfn[fy],dfn[y],1,c);
else
{
if (ans2.num == 0) ans2 = seg.query(dfn[fy],dfn[y],1,'R');//如果是从终点开始合并
else ans2.merge(seg.query(dfn[fy],dfn[y],1,'R'),ans2);
}
y = fa[fy];
fy = top[y];
}
}
if (x == y)//如果两条路径相交在同一个点
{
if (c != -1) return 0;
if (ans1.num == 0) return ans2.num;
else if (ans2.num == 0) return ans1.num;
ans1.merge(ans1,ans2);
return ans1.num;
}
if (d[x] < d[y])//y在下面
{
if (c != -1)
{
seg.modify(dfn[x] + 1,dfn[y],1,c);
return 0;
}
if (ans1.num == 0) ans1 = seg.query(dfn[x] + 1,dfn[y],1,'R');
else ans1.merge(ans1,seg.query(dfn[x] + 1,dfn[y],1,'R'));
if (ans2.num == 0) return ans1.num;
else ans1.merge(ans1,ans2);
}
else//x在下面
{
if (c != -1)
{
seg.modify(dfn[y] + 1,dfn[x],1,c);
return 0;
}
if (ans1.num == 0) ans1 = seg.query(dfn[y] + 1,dfn[x],1,'L');
else ans1.merge(ans1,seg.query(dfn[y] + 1,dfn[x],1,'L'));
if (ans2.num != 0) ans1.merge(ans1,ans2);
}
return ans1.num;
}
int main()
{
int n,p;
while (scanf("%d%d",&n,&p) != EOF)
{
for (int i = 1,u,v,t;i<=n-1;i++)
{
scanf("%d%d%d",&u,&v,&t);
w[make_pair(u,v)] = t;
w[make_pair(v,u)] = t;
g[u].push_back(v);
g[v].push_back(u);
}
dfs1(1);
dfs2(1,1);
char c[10];
seg.build(1,n,1);
while (p--)
{
scanf("%s",&c);
if (c[0] == 'Q')
{
int u,v;
scanf("%d%d",&u,&v);
if (u == v) printf("0\n");
else printf("%d\n",mapping(u,v,-1));
}
else if (c[0] == 'C')
{
int u,v,t;
scanf("%d%d%d",&u,&v,&t);
int temptemp = mapping(u,v,t);
}
}
for (int i = 1;i<=n;i++) g[i].clear();
mem(son); mem(size1); mem(d); mem(fa); mem(dfn); mem(top); mem(aa);
num = 0;
w.clear();
}
}
长链剖分
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 10;
struct node{
int u,v,nx;
}edge[MAXN<<1];
int tot,head[MAXN];
inline void add(int u,int v)
{
edge[++tot].v = v;
edge[tot].nx = head[u];
head[u] = tot;
}
int len[MAXN],son[MAXN],fa[MAXN],num;
void dfs1(int x)
{
len[x] = 1;
for (int i = head[x];i;i = edge[i].nx)
{
int v = edge[i].v;
if (v == fa[x]) continue;
fa[v] = x;
dfs1(v);
len[x] = max(len[x],len[v] + 1);
if (len[v] > len[son[x]]) son[x] = v;
}
}
int *dp[MAXN],tmp[MAXN],*id = tmp;
int ans[MAXN];
void dfs2(int x)
{
dp[x][0] = 1;
if (son[x])
{
dp[son[x]] = dp[x] + 1;
dfs2(son[x]);
ans[x] = ans[son[x]] + 1;
}
for (int i = head[x];i;i = edge[i].nx)
{
int v = edge[i].v;
if (v == fa[x] || v == son[x]) continue;
dp[v] = id;
id += len[v];
dfs2(v);
for (int j = 0;j<len[v];j++)
{
dp[x][j + 1] += dp[v][j];
if (dp[x][j+1] > dp[x][ans[x]] || (dp[x][j+1] == dp[x][ans[x]] && j + 1 < ans[x]))
ans[x] = j + 1;
}
}
if (dp[x][ans[x]] == 1) ans[x] = 0;
}
int main()
{
int n;
cin>>n;
for (int i = 1,u,v;i<n;i++)
{
scanf("%d%d",&u,&v);
add(u,v);
add(v,u);
}
dfs1(1);
dp[1] = id; id += len[1];
dfs2(1);
for (int i = 1;i<=n;i++) printf("%d\n",ans[i]);
}
LCA
树上倍增求LCA
#include<vector>
#include<cstring>
#include<stdio.h>
using namespace std;
#define mem(a) memset(a,0,sizeof(a))
const int Vertex_MAXN = 1e4 + 10;
const int DEEP = 15;
vector<int> g[Vertex_MAXN];
int up[Vertex_MAXN][DEEP] = {},d[Vertex_MAXN] = {};
void dfs(int x)
{
d[x] = d[up[x][0]] + 1;
for (int i = 1;i<=DEEP-1;i++) up[x][i] = up[up[x][i-1]][i-1];
for (int i = 0;i<g[x].size();i++)
if (g[x][i] != up[x][0])
{
up[g[x][i]][0] = x;
dfs(g[x][i]);
}
}
int LCA(int x,int y)
{
if (d[x] < d[y]) swap(x,y);
int dis = d[x] - d[y];
for (int i = DEEP-1;i>=0;i--)
if (dis & (1<<i)) x = up[x][i];
if (x == y) return x;
for (int i = DEEP-1;i>=0;i--)
{
if (up[x][i] != up[y][i])
{
x = up[x][i],y = up[y][i];
}
}
return up[x][0];
}
int main()
{
int n,in[11000] = {};
scanf("%d",&n);
for (int i = 1,u,v;i<=n-1;i++)
{
scanf("%d%d",&u,&v);//规定前者为后者的父节点
g[u].push_back(v);
in[v]++;
}
int u,v;
for (int i = 1;i<=n-1;i++)
{
if (in[i] == 0)
{
dfs(i);
break;
}
}
scanf("%d%d",&u,&v);//询问u和v的LCA
printf("%d",LCA(u,v));
mem(up); mem(d);
for (int i = 1;i<=n-1;i++) g[i].clear();
return 0;
}
线段树
树状数组
树状数组维护区间和
//树状数组维护区间和
#include<iostream>
#define ll long long
using namespace std;
int lowbit(int x)
{
return x&-x;
}
int n;
ll a[50001] = {},b[50001] = {};
void change(ll x,ll w)
{
while (x <= n)
{
b[x] += w;
x +=lowbit(x);
}
}
ll sum(int x)
{
ll num = 0;
while (x > 0)
{
num+=b[x];
x -=lowbit(x);
}
return num;
}
int main()
{
scanf("%d",&n);
for (int i = 1;i<=n;i++)
{
scanf("%d",&a[i]);
a[i] += a[i-1];
b[i] = a[i] - a[i - lowbit(i)];
}
char c[10] = {};
scanf("%s",c);
while (c[0] != 'E')
{
//E结束 Query询问x~y的区间和 Add在x处+y Sub在x处-y
int i,j;
scanf("%d%d",&i,&j);
if (c[0] == 'Q')
{
ll t = sum(j) - sum(i-1) ;
printf("%lld\n",t);
}
else if (c[0] == 'A') change(i,j);
else change(i,-j);
scanf("%s",c);
}
}
树状数组维护区间最大值
//树状数组维护区间最大值（不可有修改操作）
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int MAXN = 3e6 + 10;
#define ll long long
ll treemax[MAXN],a[MAXN];
int n;
inline int lowbit(int x)
{
return x & -x;
}
void insert(ll v,int x)
{
while (x <= n)
{
treemax[x] = max(v,treemax[x]);
x += lowbit(x);
}
}
ll query(int l,int r)
{
ll res = -9e18;
while (r >= l)
{
if (r - lowbit(r) < l)
{
res = max(res,a[r]);
r--;
}
else
{
res = max(res,treemax[r]);
r -=lowbit(r);
}
}
return res;
}
int main()
{
int m;
scanf("%d%d",&n,&m);
fill(treemax,treemax+MAXN,-9e18);
for (int i = 1;i<=n;i++)
{
scanf("%lld",&a[i]);
insert(a[i],i);
}
int l,r;
while (m--)
{
scanf("%d%d",&l,&r);
printf("%lld\n",query(l,r));
}
return 0;
}
zkw线段树
//zkw单点修改，区间查询
#include<iostream>
#include<stdio.h>
#include<cstring>
#include<math.h>
using namespace std;
const int NUM = 50000;
int m,tree[NUM * 4];
void modify(int n,int v)
{
for (tree[n += m] += v,n>>=1;n;n>>=1) tree[n] = tree[n * 2] + tree[n * 2 + 1];
}
int query(int st,int ed)
{
int ans = 0;
for (int l = m + st - 1,r = m + ed + 1;l ^ r ^ 1;l>>=1,r>>=1)
{
if (~l & 1) ans += tree[l ^ 1];
if (r & 1) ans += tree[r ^ 1];
}
return ans;
}
int main()
{
memset(tree,0,sizeof(tree));
int n;
cin>>n;
m = log(n) / log(2);
if (pow(2,m) < n) m++;
m = pow(2,m);
for (int i = m + 1;i<=m + n;i++) scanf("%d",&tree[i]);
for (int i = m - 1;i;i--) tree[i] = tree[i * 2] + tree[i * 2 + 1];
char c[10];
while (1)
{
//E结束 Query询问x~y的区间和 Add在x处+y Sub在x处-y
int x,y;
scanf("%s",c);
if (c[0] == 'E') break;
scanf("%d%d",&x,&y);
if (c[0] == 'Q') printf("%d\n",query(x,y));
else if (c[0] == 'A') modify(x,y);
else modify(x,-y);
}
return 0;
}
主席树
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 10;
struct node{
int u,v,nx;
}edge[MAXN<<1];
int tot,head[MAXN];
inline void add(int u,int v)
{
edge[++tot].v = v;
edge[tot].nx = head[u];
head[u] = tot;
}
int len[MAXN],son[MAXN],fa[MAXN],num;
void dfs1(int x)
{
len[x] = 1;
for (int i = head[x];i;i = edge[i].nx)
{
int v = edge[i].v;
if (v == fa[x]) continue;
fa[v] = x;
dfs1(v);
len[x] = max(len[x],len[v] + 1);
if (len[v] > len[son[x]]) son[x] = v;
}
}
int *dp[MAXN],tmp[MAXN],*id = tmp;
int ans[MAXN];
void dfs2(int x)
{
dp[x][0] = 1;
if (son[x])
{
dp[son[x]] = dp[x] + 1;
dfs2(son[x]);
ans[x] = ans[son[x]] + 1;
}
for (int i = head[x];i;i = edge[i].nx)
{
int v = edge[i].v;
if (v == fa[x] || v == son[x]) continue;
dp[v] = id;
id += len[v];
dfs2(v);
for (int j = 0;j<len[v];j++)
{
dp[x][j + 1] += dp[v][j];
if (dp[x][j+1] > dp[x][ans[x]] || (dp[x][j+1] == dp[x][ans[x]] && j + 1 < ans[x]))
ans[x] = j + 1;
}
}
if (dp[x][ans[x]] == 1) ans[x] = 0;
}
int main()
{
int n;
cin>>n;
for (int i = 1,u,v;i<n;i++)
{
scanf("%d%d",&u,&v);
add(u,v);
add(v,u);
}
dfs1(1);
dp[1] = id; id += len[1];
dfs2(1);
for (int i = 1;i<=n;i++) printf("%d\n",ans[i]);
}
数论
素数
素数筛
#include<iostream>
#include<math.h>
using namespace std;
const int MAX_N = 1000000;
int vis[MAX_N] = {},prime[MAX_N] = {},len = 0;
void init(int n)
{
for (int i = 2;i<=n;i++)
{
if (vis[i] == 0) prime[++len] = i;
for (int j = 1;j<=len;j++)
{
if (prime[j] * i > n) break;
vis[prime[j] * i] = 1;
if (i % prime[j] == 0) break;
}
}
}
int main()
{
init(MAX_N);
return 0;
}
图论
Tarjan
tarjan求环的个数和点的染色
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 2e2 + 10;
vector<int> g[MAXN];
int dfn[MAXN],low[MAXN],s[MAXN],vis[MAXN],num,slen,scnt,col[MAXN];
void tarjan(int u)
{
low[u] = dfn[u] = ++num;
//low存u的子树里所能到达的dfn最小的点
s[++slen] = u;//s为栈
vis[u] = 1;//标记u已经放到了栈里
for (int i = 0;i<g[u].size();i++)
{
int v = g[u][i];
if (!dfn[v])//如果v没有访问过
{
tarjan(v);
low[u] = min(low[u],low[v]);
}
else if (vis[v]) low[u] = min(low[u],dfn[v]);
//一旦遇到已入栈的点，就将该点作为连通量的根
//这里用dfn[e[i].v]更新的原因是：这个点可能
//已经在另一个强连通分量中了但暂时尚未出栈
//所以now不一定能到达low[e[i].v]但一定能到达
//dfn[e[i].v].
}
if (dfn[u] == low[u])
{
scnt++;//环的数量
do
{
vis[s[slen]] = 0;//出栈
col[s[slen]] = scnt;//染色
}while (s[slen--] != u);
}
}
int main()
{
int p;
cin>>p;
while (p--)
{
int n,m;
scanf("%d%d",&n,&m);
for (int i = 1,u,v;i<=m;i++)
{
scanf("%d%d",&u,&v);
g[u].push_back(v);
}
for (int i = 0;i<n;i++)
if (dfn[i] == 0) tarjan(i);
for (int i = 0;i<n;i++) g[i].clear();
printf("%d\n",scnt);
memset(dfn,0,sizeof(dfn));
memset(low,0,sizeof(low));
memset(s,0,sizeof(s));
memset(vis,0,sizeof(vis));
scnt = num = 0;
}
}
tarjan求无向图割边和割点
//tarjan求无向图割边和割点
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 2e4 + 10;
vector<int> g[MAXN];
int dfn[MAXN],low[MAXN],num,fa[MAXN];
struct EDGE{
EDGE(int a = 0,int b = 0):u(a),v(b){}
int u,v;
};
vector<EDGE> cutedge;
vector<int> cutnode;
bool cmp(EDGE a,EDGE b) {return a.u == b.u?a.v<b.v:a.u<b.u;}
void tarjan(int u)
{
dfn[u] = low[u] = ++num;
bool flag = false;
int son = 0;
for (int i = 0;i<g[u].size();i++)
{
int v = g[u][i];
if (v == fa[u]) continue;
if (!dfn[v])
{
son++;
fa[v] = u;
tarjan(v);
low[u] = min(low[u],low[v]);
if (low[v] >= dfn[u]) flag = true;
//判断是否存在子节点只能通过u访问到u的祖先
if (low[v] > dfn[u]) cutedge.push_back(EDGE(min(u,v),max(v,u)));
//判断割边
}
else low[u] = min(low[u],dfn[v]);
}
if ((fa[u] == 0 && son >= 2) || (fa[u] != 0 && flag)) cutnode.push_back(u);
//判断割点
//u是根节点且u有两个连通分量则u是割点
//u不是根节点，u存在一个子节点只能通过u访问到u的祖先
}
int main()
{
int n,m;
scanf("%d%d",&n,&m);
for (int i = 1,u,v;i<=m;i++)
{
scanf("%d%d",&u,&v);
g[u].push_back(v);
g[v].push_back(u);
}
for (int i = 1;i<=n;i++)
if (dfn[i] == 0) tarjan(i);
sort(cutedge.begin(),cutedge.end(),cmp);
sort(cutnode.begin(),cutnode.end());
//割点从小到大输出
//割边(u,v)，u<v，按照u为第一关键字，v为第二关键字排序
if (cutnode.size() == 0) printf("Null\n");
else
{
printf("%d",cutnode[0]);
for (int i = 1;i<cutnode.size();i++) printf(" %d",cutnode[i]);
}
puts("");
for (int i = 0;i<cutedge.size();i++)
{
printf("%d %d\n",cutedge[i].u,cutedge[i].v);
}
return 0;
}
本文来自博客园，作者：Un-Defined，转载请保留本文署名Un-Defined，并在文章顶部注明原文链接：https://www.cnblogs.com/EIPsilly/p/15728208.html
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