Going from u to v or from v to u?
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 14566 | Accepted: 3846 |
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 题目意思：给出一个结点数目为n,边数为m的有向图，若随意选出两个结点x、y，若能从x到y或者(注意，不是而且)y到x，输出Yes，否则输出No。 思路：在强连通里肯定成立，若不成立，肯定在两个强连通分量里面，那么狠容易想到缩点分块了。画了几个图后发现若分块后的图为单链那么肯定成立，若不是单链即有分叉的话那么不成立，自然而然想到拓扑了，每次删除一个入度为0的点，就看一下入度为0的点为多少个，若>1则不成立。 代码：
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 #include <iostream>
5 #include <vector>
6 #include <queue>
7 #include <stack>
8 using namespace std;
9
10 #define N 10005
11
12 int low[N], dfn[N], dfn_clock;
13 int instack[N], a[N], in[N];
14 int n, m;
15 stack<int>st;
16 vector<int>ve[N],out[N];
17 int ans, num;
18
19 void init(){
20 int i;
21 for(i=0;i<=n;i++){
22 ve[i].clear();instack[i]=1;
23 out[i].clear();
24 }
25 memset(dfn,0,sizeof(dfn));
26 memset(in,0,sizeof(in));
27 }
28
29 void tarjan(int u){
30 int v, i, j;
31 dfn[u]=low[u]=dfn_clock++;
32 st.push(u);
33 for(i=0;i<ve[u].size();i++){
34 v=ve[u][i];
35 if(!dfn[v]){
36 tarjan(v);
37 low[u]=min(low[u],low[v]);
38 }
39 else if(instack[v]){
40 low[u]=min(low[u],dfn[v]);
41 }
42 }
43 if(low[u]==dfn[u]){
44 ans++;
45 while(1){
46 v=st.top();st.pop();
47 instack[v]=0;
48 a[v]=ans;
49 if(v==u){
50 break;
51 }
52 }
53 }
54 }
55
56 main()
57 {
58 int i, j, k;
59 int x, y;
60 int t;
61 cin>>t;
62
63 while(t--){
64 scanf("%d %d",&n,&m);
65 init();
66 while(m--){
67 scanf("%d %d",&x,&y);
68 ve[x].push_back(y);
69 }
70 dfn_clock=1;ans=0;
71 for(i=1;i<=n;i++){
72 if(!dfn[i])
73 tarjan(i);
74 }
75 for(i=1;i<=n;i++){
76 for(j=0;j<ve[i].size();j++){
77 x=ve[i][j];
78 if(a[x]!=a[i]){
79 in[a[x]]++;
80 out[a[i]].push_back(a[x]);
81 }
82 }
83 }
84 int aa, f;
85 // printf("%d\n",ans);
86 // for(i=1;i<=n;i++){
87 // printf("%d ",a[i]);
88 // }
89 // cout<<endl;
90 for(i=1;i<=ans;i++){
91 aa=0;f=0;
92 for(j=1;j<=ans;j++){
93 if(!in[j]) aa++;
94 if(!in[j]&&!f){
95 for(k=0;k<out[j].size();k++){
96 in[out[j][k]]--;
97 }
98 in[j]=-1;f=1;
99 }
100
101 }
102
103 if(aa>1) break;
104
105 }
106 if(aa<=1) printf("Yes\n");
107 else printf("No\n");
108 }
109 }