POJ 2762 Going from u to v or from v to u?

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

13 31 22 33 1

Sample Output

Yes

Source

POJ Monthly--2006.02.26,zgl & twb

卡在读题。。。

题意：

给定一副有向图，选择两个点v和u，要求v能到达u或者u能到达v。

问是否可以对于图中的每一个点对<v, u>都能满足条件？输出Yes或No。

首先，想到tarjan缩点，之后原图变成DAG，通过观察发现只有当他是一条链时，才成立，dfs一遍就好了。

#include<algorithm>#include<iostream>#include<cstring>#include<cstdio>#include<stack>using namespace std;const int N=10005;const int M=60005;int T,n,m,cnt,tim,dcnt,hd[N],low[N],dfn[N],belong[N],x[M],y[M],du[N];bool flg,ins[N];stack<int>stk;struct edge{int to,nxt;}v[M];void addedge(int x,int y){v[++cnt].to=y;v[cnt].nxt=hd[x];hd[x]=cnt;}void tarjan(int u){dfn[u]=low[u]=++tim;stk.push(u);ins[u]=1;for(int i=hd[u];i;i=v[i].nxt)if(!dfn[v[i].to]){tarjan(v[i].to);low[u]=min(low[u],low[v[i].to]);}else if(ins[v[i].to])low[u]=min(low[u],dfn[v[i].to]);if(dfn[u]==low[u]){++dcnt;while(1){int t=stk.top();stk.pop();ins[t]=0;belong[t]=dcnt;if(u==t)break;}}}void dfs(int u){int sum=0;for(int i=hd[u];i;i=v[i].nxt){sum++;dfs(v[i].to);}if(sum>1)flg=1;}int main(){scanf("%d",&T);while(T--){flg=0;cnt=dcnt=tim=0;memset(hd,0,sizeof(hd));memset(du,0,sizeof(du));memset(dfn,0,sizeof(dfn));memset(low,0,sizeof(low));scanf("%d%d",&n,&m);for(int i=1;i<=m;i++){scanf("%d%d",&x[i],&y[i]);addedge(x[i],y[i]);}for(int i=1;i<=n;i++)if(!dfn[i])tarjan(i);cnt=0;memset(hd,0,sizeof(hd));for(int i=1;i<=m;i++)if(belong[x[i]]!=belong[y[i]]){addedge(belong[x[i]],belong[y[i]]);du[belong[y[i]]]++;}int sum=0;for(int i=1;i<=dcnt;i++)if(du[i]==0){sum++;dfs(i);}if(sum>1||flg)printf("No\n");elseprintf("Yes\n");}return 0;}