学渣乱搞系列之Tarjan模板合集

时间:2024-01-12 21:53:02

学渣乱搞系列之Tarjan模板合集

            by 狂徒归来

一、求强连通子图

 #include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = ;
int dfn[maxn],low[maxn],belong[maxn];
bool instack[maxn];
vector<int>g[maxn];
stack<int>stk;
int n,m,cnt,scc;
void tarjan(int u) {
dfn[u] = low[u] = ++cnt;
stk.push(u);
instack[u] = true;
for(int i = ; i < g[u].size(); i++) {
if(!dfn[g[u][i]]) {
tarjan(g[u][i]);
low[u] = min(low[u],low[g[u][i]]);
} else if(instack[g[u][i]]) low[u] = min(low[u],dfn[g[u][i]]);
}
if(dfn[u] == low[u]) {
scc++;
int v;
do {
v = stk.top();
instack[v] = false;
belong[v] = scc;
stk.pop();
} while(v != u);//每一个强连通块内de点,以及其属于的scc
}
}
int main() {
int i,j,u,v,a,b;
while(~scanf("%d %d",&n,&m)) {
for(i = ; i <= n; i++) {
dfn[i] = belong[i] = ;
instack[i] = false;
g[i].clear();
}
cnt = scc = ;
while(!stk.empty()) stk.pop();
for(i = ; i <= m; i++) {
scanf("%d %d",&u,&v);
g[u].push_back(v);
}
for(i = ; i <= n; i++)
if(!dfn[i]) tarjan(i);
}
return ;
}

二、求割点

 const int maxn = ;
vector<int>g[maxn];
bool iscut[maxn];
int dfn[maxn],low[maxn],cnt,vis[maxn];
void tarjan(int u,int fa) {
dfn[u] = low[u] = ++cnt;
vis[u] = ;
int son = ;
for(int i = ; i < g[u].size(); i++) {
if(!vis[g[u][i]]) {
tarjan(g[u][i],u);
son++;
low[u] = min(low[u],low[g[u][i]]);
if(fa == - && son > || fa != - && low[g[u][i]] >= dfn[u])
iscut[u] = true;
} else if(vis[g[u][i]] == ) low[u] = min(low[u],dfn[g[u][i]]);
}
vis[u] = ;
}

三、求割边/桥

 int ret;
void tarjan(int u,int fa) {
dfn[u] = low[u] = ++clk;
bool flag = false;
for(int i = head[u]; ~i; i = e[i].next) {
if(!flag && e[i].to == fa) {
flag = true;
continue;
}
if(!dfn[e[i].to]) {
tarjan(e[i].to,u);
low[u] = min(low[u],low[e[i].to]);
if(low[e[i].to] > dfn[u]) {
e[i].cut = e[i^].cut = true;
++ret;
}
} else low[u] = min(low[u],dfn[e[i].to]);
}
}

四、求LCA