hdu1269（有向图强连通分量）

hdu1269

题意

判断对于任意两点是否都可以互相到达（判断有向图强连通分量个数是否为 1 ）。

分析

Tarjan 算法实现。

code

#include<bits/stdc++.h>
typedef long long ll;
using namespace std;
const int MAXN = 2e5 + 10;
int n, m;
struct Edge {
    int to, next;
}e[MAXN];
int cnt, head[MAXN];
void addedge(int u, int v) {
    e[cnt].to = v;
    e[cnt].next = head[u];
    head[u] = cnt++;
}
int sz, dfn[MAXN], low[MAXN], vis[MAXN];
int scc;
stack<int> sta;
void tarjan(int u) {
    dfn[u] = low[u] = ++sz;
    vis[u] = 1;
    sta.push(u);
    for(int i = head[u]; ~i; i = e[i].next) {
        int v = e[i].to;
        if(!dfn[v]) {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        } else if(vis[v] && low[u] > dfn[v]) {
            low[u] = dfn[v];
        }
    }
    if(low[u] == dfn[u]) {
        scc++;
        while(1) {
            int id = sta.top(); sta.pop();
            vis[id] = 0;
            if(id == u) break;
        }
    }
}
int main() {
    while(scanf("%d%d", &n, &m) && (n + m)) {
        memset(vis, 0, sizeof vis);
        while(!sta.empty()) sta.pop();
        scc = cnt = sz = 0;
        memset(head, -1, sizeof head);
        memset(dfn, 0, sizeof dfn);
        for(int i = 0; i < m; i++) {
            int u, v;
            scanf("%d%d", &u, &v);
            addedge(u, v);
        }
        for(int i = 1; i <= n; i++) {
            if(!dfn[i]) tarjan(i);
        }
        puts(scc == 1 ? "Yes" : "No");
    }
    return 0;
}