POJ 2672 Tarjan + 缩点 + 拓扑思想

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.

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

来源： http://poj.org/problem?id=2762

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120

#include<cstdio> #include<cstring> #include<iostream> #include<string> #include<vector> #include<stack> #include<set> #include<algorithm> using namespace std; #define N 1002 vector<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\n");             return ;         }         inDegree[pos ] = -1;         for(int i=1;i<=cnt;i++)         {             if(map[pos][i] == 1)                 inDegree[i]--;         }     }     printf("Yes\n"); }   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\n");             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\n",i,inDegree[i]);         tsort();     } } 来源： http://tool.oschina.net/highlight