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HDOJ 1269 迷宫城堡

2014-10-08 09:26 246 查看
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 IDSubmit TimeJudge StatusPro.IDExe.TimeExe.MemoryCode Len.LanguageAuthor
118223082014-10-08 09:19:41Accepted126978MS1848K1169 BG++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;
}
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