HDOJ 1269 迷宫城堡

题意：判断有向图中是否存在任意点n到各点都为连通

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

思路：强连通分量，tarjan，判断强连通分量的数量，直接套模板了。

以下为AC代码：

Run ID	Submit Time	Judge Status	Pro.ID	Exe.Time	Exe.Memory	Code Len.	Language	Author
11822308	2014-10-08 09:19:41	Accepted	1269	78MS	1848K	1169 B	G++	luminous11

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<stack>

using namespace std;

#define MAX_V 11111

int V, E;
int dfn[MAX_V], low[MAX_V];
bool instack[MAX_V];
int din;
int cnt;
stack<int> s;
vector<int> Edge[MAX_V];

void tarjan ( int x )
{
    instack[x] = true;
    dfn[x] = low[x] = din++;
    s.push ( x );
    for ( int i = 0; i < Edge[x].size(); i ++ )
    {
        int j = Edge[x][i];
        if ( ! dfn[j] )
        {
            tarjan ( j );
            low[x] = min ( low[x], low[j] );
        }
        else if (instack[j])
            low[x] = min ( low[x], dfn[j] );
    }
    if ( dfn[x] == low[x] )
    {
        cnt++;
        int tmp;
        do
        {
            tmp = s.top();
            s.pop();
            instack[tmp] = false;
        }
        while ( x != tmp );
    }
}

int main()
{
    while ( scanf ( "%d%d" , &V, &E ) && ( V || E ) )
    {
        for ( int i = 0; i < V; i ++ )
            Edge[i].clear();
        for ( int i = 0; i < E; i ++ )
        {
            int u, v;
            scanf ( "%d%d", &u, &v);
            Edge[u - 1].push_back(v - 1);
        }
        memset ( dfn, 0, sizeof ( dfn ) );
        memset ( instack, false, sizeof ( instack ) );
        din = 0;
        cnt = 0;
        for ( int i = 0; i < V; i ++ )
        {
            if ( ! dfn[i] )
                tarjan ( i );
        }
        if ( cnt == 1 )
            printf ( "Yes\n" );
        else
            printf ( "No\n" );
    }
    return 0;
}