Going from u to v or from v to u?
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 17383 | Accepted: 4660 |
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.
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
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#include<cstdio>#include<cstring>#include<iostream>#include<string>#include<vector>#include<stack>#include<set>#include<algorithm>using namespace std;#define N 1002vector<int>Gra[N];stack<int>Sta;int map[N][N];int dfn[N],low[N],inStack[N],belong[N],Time,cnt;int inDegree[N];void init(){ Time = cnt = 0; memset(dfn,0,sizeof(dfn)); memset(low,0,sizeof(dfn)); memset(inStack,0,sizeof(inStack)); memset(inDegree,0,sizeof(inDegree)); memset(belong,0,sizeof(belong)); for(int i=0;i<N;i++) Gra[i].clear(); memset(map,0,sizeof(map)); while(!Sta.empty()) Sta.pop();}void Tarjan(int s){ dfn[s] = low[s] = ++Time; inStack[s] = 1; Sta.push(s); for(int i=0;i<Gra[s].size();i++) { int j = Gra[s][i]; if(dfn[j] == 0){ Tarjan(j); low[s] = min(low[s], low[j]); } else if(inStack[j] == 1){ low[s] = min(low[s], dfn[j]); } } if(dfn[s] == low[s]) { cnt ++; while(!Sta.empty()){ int temp = Sta.top(); Sta.pop(); inStack[temp] = 0; belong[temp] = cnt; if(temp == s) break; } } return;}void tsort(){ for(int k=0;k<cnt;k++){ int fuck = 0,pos; for(int i=1;i<=cnt;i++) { if(inDegree[i] == 0) { fuck ++; pos = i; } } if(fuck > 1){ printf("No
"); return ; } inDegree[pos ] = -1; for(int i=1;i<=cnt;i++) { if(map[pos][i] == 1) inDegree[i]--; } } printf("Yes
");}int main(){ int noc; cin>>noc; while(noc--) { init(); int n,m,x,y; scanf("%d%d",&n,&m); for(int i=0;i<m;i++) { scanf("%d%d",&x,&y); Gra[x].push_back(y); } for(int i=1;i<=n;i++) if(dfn[i] == 0) Tarjan(i); if(cnt == 1) { printf("Yes
"); continue; } for(int i=1;i<=n;i++) { for(int j=0;j<Gra[i].size();j++) { int k = Gra[i][j]; if(belong[i]!=belong[k]){ if(map[belong[i]][belong[k]] == 0){ map[belong[i]][belong[k]] = 1; inDegree[belong[k]]++; } } } } for(int i=1;i<=cnt;i++) printf("%d %d
",i,inDegree[i]); tsort(); }} |