Code[vs] 1506传话（tarjan缩点）

http://codevs.cn/problem/1506/

这道题就是问每个点是否存在于一个环里面，所以用tarjan跑一遍，然后检查他所在的强联通分量里面是否存在多于一个的变量数目。若变量数目大于1，输出T，否则输出F

代码如下：

#include<bits/stdc++.h>using namespace std;int n, m;vector <int> G[1005], com[1005];int dfn[1005], low[1005], key[1005];int ind, com_num;int in[1005];stack <int> s;void tarjan(int x) {key[x] = 2;dfn[x] = low[x] = ind++;s.push(x);for(int i = 0; i < G[x].size(); i++) {int v = G[x][i];if(dfn[v] == -1) {tarjan(v);dfn[x] = min(dfn[x], dfn[v]);} else if(key[v] == 2) {dfn[x] = min(dfn[x], low[v]);}}if(dfn[x] == low[x]) {while(1) {int t = s.top();s.pop();in[t] = com_num;com[com_num].push_back(t);key[t] = 1;if(t == x)break;}com_num++;}}int main() {int a, b;memset(dfn, -1, sizeof(dfn));memset(low, -1, sizeof(low));int n, m;cin >> n >> m;while(m--) {scanf("%d%d", &a, &b);G[a].push_back(b);}for(int i = 1; i <= n; i++)if(dfn[i] == -1)tarjan(i);for(int i = 1; i <= n; i++) {if(com[in[i]].size() > 1)cout << "T" << endl;elsecout << "F" << endl;}return 0;}