[HDU 1269] 迷宫城堡 强连通分量

http://acm.hdu.edu.cn/showproblem.php?pid=1269

题意：中文题。。

思路：Tarjan算法，赤果果的强连通分量，直接上模板

#include 
#include 
#include 
#include 
#include 

using namespace std;

const int maxn = 10005;

stack<int> sta;
vector<int> side[100005];

int time, sum;
int low[maxn], dfn[maxn];
bool instack[maxn], vis[maxn];

int Tarjan(int rt)
{
    sta.push(rt);
    vis[rt] = true;
    instack[rt] = true;
    low[rt] = dfn[rt] = ++time;
    int len = side[rt].size();
    for(int i = 0; i < len; i++)
    {
        int rw = side[rt][i];
        if(!vis[rw]){
            Tarjan(rw);
            low[rt] = min(low[rt], low[rw]);
        }
        else if(instack[rw]){
            low[rt] = min(low[rt], dfn[rw]);
        }
    }
    if(low[rt] == dfn[rt]){
        int top;
        do{
            top =sta.top();
            sta.pop();
            instack[top] = false;
        }while(top != rt);
        sum++;
    }
    return 0;
}

int main()
{
    int n, m;
    while(~scanf("%d%d", &n, &m) && (n || m)){
        for(int i = 0; i <= n; i++){
            side[i].clear();
        }
        int x, y;
        for(int i = 0; i < m; i++){
            scanf("%d%d", &x, &y);
            side[x].push_back(y);
        }
        sum = time = 0;
        memset(low, 0, sizeof(low));
        memset(dfn, 0, sizeof(dfn));
        memset(vis, false, sizeof(vis));
        memset(instack, false, sizeof(instack));
        for(int i = 1; i <= n; i++){
            if(!dfn[i])
                Tarjan(1);
        }
        if(sum == 1){
            printf("Yes\n");
        }
        else{
            printf("No\n");
        }
    }
}