迷宮城堡
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 16123 Accepted Submission(s): 7088
原題鏈接:http://acm.hdu.edu.cn/showproblem.php?pid=1269
題意:判斷圖的強連通分量是否爲一.模板題.
Targan算法介紹及模板:http://blog.csdn.net/hurmishine/article/details/75248876
AC代碼:
/**
* 行有餘力,則來刷題!
* 博客鏈接:http://blog.csdn.net/hurmishine
* 個人博客網站:http://wuyunfeng.cn/
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <stack>
#include <algorithm>
using namespace std;
const int maxn=10000+5;
vector<int>G[maxn];
int n,m;
int index;
int cnt;
int low[maxn],dfn[maxn];
bool vis[maxn];//是否在棧裏
void Init()
{
cnt=index=0;
for(int i=0;i<=n;i++)
low[i]=dfn[i]=0;
}
void Tarjan(int u)
{
stack<int>s;
s.push(u);
vis[u]=true;
low[u]=dfn[u]=++index;
for(int i=0;i<G[u].size();i++)
{
int v = G[u][i];
if(!dfn[v])
{
Tarjan(v);
low[u] = min(low[u],low[v]);
}
else if(vis[v])
{
low[u] = min(low[u],dfn[v]);
}
}
if(low[u] == dfn[u])
{
cnt++;
int x;
do
{
x = s.top();
s.pop();
vis[x]=false;
}while(x!=u);
}
}
int main()
{
//freopen("C:\\Documents and Settings\\Administrator\\桌面\\data.txt","r",stdin);
while(cin>>n>>m,n+m)
{
for(int i=0;i<=n;i++)
G[i].clear();
int x,y;
while(m--)
{
scanf("%d%d",&x,&y);
G[x].push_back(y);
}
Init();
for(int i=1;i<=n;i++)
{
if(!dfn[i])
Tarjan(i);
}
if(cnt == 1)
cout<<"Yes"<<endl;
else
cout<<"No"<<endl;
}
return 0;
}