11175-From D to E and Back（思维）

Problem UVA11175-From D to E and Back

Accept: 164  Submit: 607
Time Limit: 3000 mSec

Problem Description

Take any directed graph D with n vertices and m edges. You can make the Lying graph E of B in the following way. E will have m vertices, one for each edge of D. For example, if D has an edge uv, then E will have a vertex called uv. Now, whenever D has edges uv and vw, E will have an edge from vertex uv to vertex vw. There are no other edges in E. You will be given a graph E and will have to determine whether it is possible for E to be the Lying graph of some directed graph D.

Input

The first line of input gives the number of cases, N (N < 220). N test cases follow. Each one starts with two lines containing m (0 ≤ m ≤ 300) and k. The next k lines will each contain a pair of vertices, x and y, meaning that there is an edge from x to y in E. The vertices are numbered from 0 to m−1

 Output

For each test case, output one line containing ‘Case #x:’ followed by either ‘Yes’ or ‘No’, depending on whether E is a valid Lying graph or not. Note that D is allowed to have duplicate edges and self-edges.

 

 Sample Input

4 2 1 0 1 5 0 4 3 0 1 2 1 2 3 3 9 0 1 0 2 1 2 1 0 2 0 2 1 0 0 1 1 2 2

 

Sample Output

Case #1: Yes

Case #2: Yes

Case #3: No

Case #4: Yes

题解：这种结论题真的是做不来。首先第一感觉是这怎么会不存在呢，然后样例三强势打脸，但是感觉上不成立的情况应该很少，但是少到何种程度完全没有认识，最后思来想去还是看了题解，题解都是千篇一律的结论，并且没有人证明那是充要的，至多证明是必要的，做这个题就当涨见识了。

 #include <bits/stdc++.h>

 using namespace std;

 const int maxn =  + ;
 int gra[maxn][maxn];

 int m, t;

 int read() {
     int q = ;
     char ch = ' ';
     while (ch<'' || ch>'') ch = getchar();
     while ('' <= ch && ch <= '') {
         q = q *  + ch - '';
         ch = getchar();
     }
     return q;
 }

 bool solve() {
     for (int i = ; i < m; i++) {
         for (int j = ; j < m; j++) {
             int f1 = , f2 = ;
             for (int k = ; k < m; k++) {
                 if (gra[i][k] && gra[j][k]) f1 = ;
                 if (gra[i][k] ^ gra[j][k]) f2 = ;
                 if (f1 && f2) return false;
             }
         }
     }
     return true;
 }

 int T = ;

 int main()
 {
     int iCase;
     iCase = read();
     while (iCase--) {
         memset(gra, , sizeof(gra));
         m = read(), t = read();
         int u, v;
         for (int i = ; i < t; i++) {
             u = read(), v = read();
             gra[u][v] = ;
         }

         printf("Case #%d: ",T++);
         if (solve()) printf("Yes\n");
         else printf("No\n");
     }
     return ;
 }