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.
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
1 #include<cstdio> 2 #include<vector> 3 #include<queue> 4 #include<stack> 5 #include<algorithm> 6 #include<cstring> 7 using namespace std; 8 9 vector<int>G[1005]; 10 vector<int>G1[1005]; 11 stack<int>s; 12 int dfn[1005],lowlink[1005],sccno[1005],in[1005]; 13 int m,n,dfs_clock,scc_cnt; 14 bool ok; 15 16 void init() 17 { 18 memset(dfn,0,sizeof(dfn)); 19 memset(lowlink,0,sizeof(lowlink)); 20 memset(sccno,0,sizeof(sccno)); 21 memset(in,0,sizeof(in)); 22 for(int i=0;i<n;i++)G[i].clear(),G1[i].clear(); 23 dfs_clock=scc_cnt=0; 24 } 25 26 void tarjan(int u) 27 { 28 lowlink[u]=dfn[u]=++dfs_clock; 29 s.push(u); 30 for(int i=0;i<G[u].size();i++) 31 { 32 int v=G[u][i]; 33 if(!dfn[v]) 34 { 35 tarjan(v); 36 lowlink[u]=min(lowlink[u],lowlink[v]); 37 } 38 else if(!sccno[v]) 39 lowlink[u]=min(lowlink[u],dfn[v]); 40 } 41 if(lowlink[u]==dfn[u]) 42 { 43 scc_cnt++; 44 while(1) 45 { 46 int x=s.top(); 47 s.pop(); 48 sccno[x]=scc_cnt; 49 if(x==u)break; 50 } 51 } 52 } 53 54 void toposort() 55 { 56 ok=1; 57 int sum=0; 58 queue<int>q; 59 for(int i=1;i<=scc_cnt;i++) 60 if(in[i]==0) 61 q.push(i); 62 while(!q.empty()) 63 { 64 if(q.size()!=1) 65 { 66 ok=0; 67 return; 68 } 69 int x=q.front(); 70 q.pop(); 71 sum++; 72 for(int i=0;i<G1[x].size();i++) 73 if(--in[G1[x][i]]==0) 74 q.push(G1[x][i]); 75 } 76 if(sum!=scc_cnt) 77 { 78 ok=0; 79 return; 80 } 81 } 82 83 int main() 84 { 85 int T; 86 scanf("%d",&T); 87 while(T--) 88 { 89 init(); 90 scanf("%d%d",&n,&m); 91 for(int i=0;i<m;i++) 92 { 93 int u,v; 94 scanf("%d%d",&u,&v); 95 G[u].push_back(v); 96 } 97 for(int i=1;i<=n;i++) 98 if(!dfn[i]) 99 tarjan(i); 100 for(int i=1;i<=n;i++) 101 for(int j=0;j<G[i].size();j++) 102 { 103 int v=G[i][j]; 104 if(sccno[i]!=sccno[v]) 105 { 106 G1[sccno[i]].push_back(sccno[v]); 107 in[sccno[v]]++; 108 } 109 } 110 toposort(); 111 ok?puts("Yes"):puts("No"); 112 } 113 return 0; 114 }