Going from u to v or from v to u?
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 16486 | Accepted: 4386 |
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
题意:
有一张有向图，然后问对于任意的2个点，是否存在x可以到达y，或者y能够到达x。
思路:
可以先强连通缩点，这样就没有环的情况。然后这样就是一张新的图。也就是相当于在新的图中判断任意2个点能否存在一条路径。这可以用topo排序来解决。如果存在2个点，他们的入度为0，说明这2个点是不能相互到达的。
/*
* Author: sweat123
* Created Time: 2016/6/25 12:33:09
* File Name: main.cpp
*/
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<string>
#include<vector>
#include<cstdio>
#include<time.h>
#include<cstring>
#include<iostream>
#include<algorithm>
#define INF 1<<30
#define MOD 1000000007
#define ll long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define pi acos(-1.0)
using namespace std;
const int MAXN = 20000;
struct node{
int from;
int to;
int next;
}edge[MAXN*1000];
int pre[MAXN],vis[MAXN],dfn[MAXN],low[MAXN],n,m,ind;
int f[MAXN],num,k;
stack<int>s;
void add(int x,int y){
edge[ind].from = x;
edge[ind].to = y;
edge[ind].next = pre[x];
pre[x] = ind ++;
}
void dfs(int rt){
vis[rt] = 1;
dfn[rt] = low[rt] = ++k;
s.push(rt);
for(int i = pre[rt]; i != -1; i = edge[i].next){
int t = edge[i].to;
if(!dfn[t]){
dfs(t);
low[rt] = min(low[rt],low[t]);
} else if(vis[t]){
low[rt] = min(low[rt],dfn[t]);
}
}
if(low[rt] == dfn[rt]){
++ num;
while(!s.empty()){
int tp = s.top();
s.pop();
vis[tp] = 0;
f[tp] = num;
if(tp == rt)break;
}
}
}
int x[MAXN],y[MAXN],ans,in[MAXN];
int ok(){
queue<int>q;
for(int i = 1; i <= num; i++){
if(in[i] == 0){
q.push(i);
}
}
if(q.size() > 1)return 0;
while(!q.empty()){
int tp = q.front();
q.pop();
for(int i = pre[tp]; i != -1; i = edge[i].next){
int t = edge[i].to;
in[t] --;
if(in[t] == 0){
q.push(t);
}
}
if(q.size() > 1)return 0;
}
return 1;
}
int main(){
int t;
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&m);
ind = 0;
while(!s.empty())s.pop();
memset(pre,-1,sizeof(pre));
for(int i = 1; i <= m; i++){
scanf("%d%d",&x[i],&y[i]);
add(x[i],y[i]);
}
k = 0;
num = 0;
memset(vis,0,sizeof(vis));
memset(dfn,0,sizeof(dfn));
memset(low,0,sizeof(low));
memset(f,0,sizeof(f));
for(int i = 1; i <= n; i++){
if(!dfn[i])dfs(i);
}
memset(pre,-1,sizeof(pre));
memset(in,0,sizeof(in));
int ret = ind;
for(int i = 0; i < ret; i++){
int u = f[edge[i].from];
int v = f[edge[i].to];
if(u != v){
add(u,v);
in[v] ++;
}
}
if(ok()){
printf("Yes\n");
} else{
printf("No\n");
}
}
return 0;
}