POJ 2762 Going from u to v or from v to u? Tarjan算法 学习例题

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 17104		Accepted: 4594

Description

In order to make their sons brave, Jiajia and Wind take them to a big cave. The cave has n rooms, and one-way corridors connecting some rooms. Each time, Wind choose two rooms x and y, and ask one of their little sons go from one to the other. The son can either go from x to y, or from y to x. Wind promised that her tasks are all possible, but she actually doesn't know how to decide if a task is possible. To make her life easier, Jiajia decided to choose a cave in which every pair of rooms is a possible task. Given a cave, can you tell Jiajia whether Wind can randomly choose two rooms without worrying about anything?

Input

The first line contains a single integer T, the number of test cases. And followed T cases. 

The first line for each case contains two integers n, m(0 < n < 1001,m < 6000), the number of rooms and corridors in the cave. The next m lines each contains two integers u and v, indicating that there is a corridor connecting room u and room v directly. 

Output

The output should contain T lines. Write 'Yes' if the cave has the property stated above, or 'No' otherwise.

Sample Input

1
3 3
1 2
2 3
3 1

Sample Output

Yes

Source

POJ Monthly--2006.02.26,zgl & twb

题目大意：n个山洞，对于每两个山洞s,e，都满足s可以到达e或者e可以到达s，则输出Yes,否则输出No。（s到e或者e到s满足一个就可以）

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<stack>
 #define maxm 6010
 #define maxn 1010
 using namespace std;
 int T,n,m;
 struct edge{
     int u,v,next;
 }e[maxm],ee[maxm];
 int head[maxn],js,headd[maxn],jss;
 bool exist[maxn];
 int visx,cur;// cur--缩出的点的数量
 int dfn[maxn],low[maxn],belong[maxn];
 int rudu[maxn],chudu[maxn];
 stack<int>st;
 void init(){
     memset(rudu,,sizeof(rudu));memset(chudu,,sizeof(chudu));
     memset(head,,sizeof(head));memset(headd,,sizeof(headd));
     jss=js=visx=cur=;
     memset(exist,false,sizeof(exist));
     while(!st.empty())st.pop();
     memset(dfn,-,sizeof(dfn));memset(low,-,sizeof(low));
     memset(belong,,sizeof(belong));
 }
 void add_edge1(int u,int v){
     e[++js].u=u;e[js].v=v;
     e[js].next=head[u];head[u]=js;
 }
 void tarjan(int u){
     dfn[u]=low[u]=++visx;
     exist[u]=true;
     st.push(u);
     for(int i=head[u];i;i=e[i].next){
         int v=e[i].v;
         if(dfn[v]==-){
             tarjan(v);
             if(low[v]<low[u]) low[u]=low[v];
         }
         else if(exist[v]&&low[u]>dfn[v]) low[u]=dfn[v];
     }
     int j;
     if(low[u]==dfn[u]){
         ++cur;
         do{
             j=st.top();st.pop();exist[j]=false;
             belong[j]=cur;
         }while(j!=u);
     }
 }
 void add_edge2(int u,int v){
     ee[++jss].u=u;ee[jss].v=v;
     ee[jss].next=headd[u];headd[u]=jss;
 }
 bool topo()
 {
     int tp=,maxt=;
     for(int i=;i<=cur;i++)
     {
         if(rudu[i]==){
             tp++;
         }
         if(chudu[i]>maxt)maxt=chudu[i];
     }
     if(tp>)return ;// 入读等于0的缩点只能有一个 否则..
     if(maxt>)return ;
     return ;
 }
 int main()
 {
     scanf("%d",&T);
     while(T--){
         scanf("%d%d",&n,&m);
         init();
         int u,v;
         for(int i=;i<m;i++){
             scanf("%d%d",&u,&v);add_edge1(u,v);
         }
         for(int i=;i<=n;i++){// 求强连通分量
             if(dfn[i]==-) tarjan(i);
         }
         for(int i=;i<=js;i++){
             int u=e[i].u,v=e[i].v;
             if(belong[u]!=belong[v]){
                 add_edge2(belong[u],belong[v]);
                 rudu[belong[v]]++;chudu[belong[u]]++;
             }
         }
         if(topo()==) printf("Yes\n");
         else printf("No\n");
     }
     return ;
 }

思路：首先建一个有向图，进行Tarjan算法，进行缩点，缩完点之后，再建一张有向图（这张图只能成一条链），对其进行检验，入度为0的店只能有一个或没有。。