Going from u to v or from v to u?
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 19552 | Accepted: 5262 |
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
Source
POJ Monthly--2006.02.26,zgl & twb
题意: 给你一个有向图,如果对于图中的任意一对点u和v都有一条从u到v的路或从v到u的路,那么就输出’Yes’,否则输出’No’.
分析:
首先求出该图的所有强连通分量，把所有分量缩点，构成一个新的DAG图。现在的问题变成了：该DAG图是否对于任意两点都存在一条路。
多画几个图可以知道该DAG图只能是一条链的时候才行(自己画图验证一下)，用拓扑排序验证(队列中始终只有一个元素)。
代码：
1 #include"bits/stdc++.h" 2 3 #define db double 4 #define ll long long 5 #define vl vector<ll> 6 #define ci(x) scanf("%d",&x) 7 #define cd(x) scanf("%lf",&x) 8 #define cl(x) scanf("%lld",&x) 9 #define pi(x) printf("%d ",x) 10 #define pd(x) printf("%f ",x) 11 #define pl(x) printf("%lld ",x) 12 #define rep(i, a, n) for (int i=a;i<n;i++) 13 #define per(i, a, n) for (int i=n-1;i>=a;i--) 14 #define fi first 15 #define se second 16 using namespace std; 17 typedef pair<int, int> pii; 18 const int N = 6e3 + 5; 19 const int mod = 1e9 + 7; 20 const int MOD = 998244353; 21 const db PI = acos(-1.0); 22 const db eps = 1e-10; 23 const int inf = 0x3f3f3f3f; 24 const ll INF = 0x3fffffffffffffff; 25 int in[N]; 26 int out[N]; 27 int head[N],h[N]; 28 int low[N],dfn[N]; 29 int ins[N],beg[N]; 30 int cnt,id,num,cntt; 31 int n, m, t; 32 queue<int> q; 33 stack<int> s; 34 struct P { 35 int to, nxt; 36 } e[2 * N], g[2 * N]; 37 38 void add(int u, int v) {//原图 39 e[cnt].to = v; 40 e[cnt].nxt = head[u]; 41 head[u] = cnt++; 42 } 43 void ADD(int u, int v) {//缩点图 44 g[cntt].to = v; 45 g[cntt].nxt = h[u]; 46 h[u] = cntt++; 47 } 48 void tarjan(int u) 49 { 50 low[u]=dfn[u]=++id; 51 ins[u]=1; 52 s.push(u); 53 for(int i=head[u];~i;i=e[i].nxt){ 54 int v=e[i].to; 55 if(!dfn[v]) tarjan(v),low[u]=min(low[u],low[v]); 56 else if(ins[v]) low[u]=min(low[u],dfn[v]); 57 } 58 if(low[u]==dfn[u]){ 59 int v; 60 do{ 61 v=s.top();s.pop(); 62 beg[v]=num; 63 ins[v]=0; 64 }while(u!=v); 65 num++; 66 } 67 } 68 bool work() 69 { 70 while(!q.empty()) q.pop(); 71 for(int i=0;i<num;i++){ 72 if(!in[i]) q.push(i); 73 // if(in[i]>1||out[i]>1) return 0;//链上点出度入度都不大于1 74 } 75 int ans=0; 76 while(!q.empty()){ 77 if(q.size()>1) return 0; 78 int u=q.front();q.pop(); 79 ans++; 80 for(int i=h[u];~i;i=g[i].nxt){ 81 int v=g[i].to; 82 in[v]--; 83 if(!in[v]) q.push(v); 84 } 85 } 86 return ans==num;//成链 87 } 88 void init() 89 { 90 cnt = num = id = cntt= 0; 91 memset(in, 0, sizeof(in)); 92 memset(out, 0, sizeof(out)); 93 memset(head, -1, sizeof(head)); 94 memset(h, -1, sizeof(h)); 95 memset(low,0, sizeof(low)); 96 memset(dfn,0, sizeof(dfn)); 97 memset(ins,0, sizeof(ins)); 98 memset(beg,0, sizeof(beg)); 99 } 100 int main() { 101 ci(t); 102 while(t--) 103 { 104 ci(n),ci(m); 105 init(); 106 for (int i = 0; i < m; i++) { 107 int x, y; 108 ci(x), ci(y); 109 add(x, y); 110 } 111 for(int i=1;i<=n;i++) if(!dfn[i]) tarjan(i); 112 for(int i=1;i<=n;i++){ 113 for(int j=head[i];~j;j=e[j].nxt){ 114 int v=e[j].to; 115 if(beg[i]!=beg[v]) out[beg[i]]++,in[beg[v]]++,ADD(beg[i],beg[v]);//缩点后的图 116 } 117 } 118 puts(work()?"Yes":"No"); 119 } 120 }