hdoj 1269 迷宫城堡(强连通分量)

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

思路分析：该问题要求判断是否每两个房间都可以相互到达，即求该有向图中的所有点是否只构成一个强连通图分量，使用Tarjan算法即可求解；

代码如下：

#include <stack>
#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

const int MAX_N =  + ;
vector<int> G[MAX_N];
stack<int> S;
int pre[MAX_N], lowlink[MAX_N], scc_no[MAX_N];
int dfs_clock, scc_cnt;

inline int Min(int a, int b) { return a < b ? a : b; }
void Dfs(int u)
{
    pre[u] = lowlink[u] = ++dfs_clock;
    S.push(u);

    for (int i = ; i < G[u].size(); ++i)
    {
        int v = G[u][i];

        if (!pre[v])
        {
            Dfs(v);
            lowlink[u] = Min(lowlink[u], lowlink[v]);
        } else if (!scc_no[v])
            lowlink[u] = Min(lowlink[u], lowlink[v]);
    }
    if (lowlink[u] == pre[u])
    {
        scc_cnt++;
        for (;;)
        {
            int x = S.top();
            S.pop();
            scc_no[x] = scc_cnt;
            if (x == u)  break;
        }
    }
}

inline void FindScc(int n)
{
    dfs_clock = scc_cnt = ;
    memset(pre, , sizeof(pre));
    memset(lowlink, , sizeof(lowlink));
    memset(scc_no, , sizeof(scc_no));
    for (int i = ; i <= n; ++i)
    if (!pre[i]) Dfs(i);
}

int main()
{
    int ver_num, road_num;

    while (scanf("%d %d", &ver_num, &road_num) != EOF
        && (ver_num + road_num))
    {
        int ver_1, ver_2;

        for (int i = ; i < MAX_N; ++i)
            G[i].clear( );
        for (int i = ; i < road_num; ++i)
        {
            scanf("%d %d", &ver_1, &ver_2);
            G[ver_1].push_back(ver_2);
        }
        FindScc(ver_num);
        while (!S.empty( ))
            S.pop( );
        if (scc_cnt > )
            printf("No\n");
        else
            printf("Yes\n");
    }
    return ;
}