POJ 3660

题目大意:有N个人，进行M场比赛，M场比赛只给两个参赛选手，第一个就是赢了的参赛选手，求排名固定的参赛选手

结题思路:如果一个人被赢了m次，赢了别人n次，如果m+n==N-1，那么满足条件,Floyd

 1 #include <iostream>
 2 #include <cstring>
 3 #include <algorithm>
 4 #define MAXVERTEXNUM 110
 5 #define INF 1000000
 6 using namespace std;
 7 
 8 int Nv, Ne;
 9 int D[MAXVERTEXNUM][MAXVERTEXNUM];
10 
11 void Floyd()
12 {
13     for (int k = 1; k <= Nv; ++k)
14         for (int i = 1; i <= Nv; ++i)
15             for (int j = 1; j <= Nv; ++j)
16                 if (D[i][j] > D[i][k] + D[k][j])
17                     D[i][j] = D[i][k] + D[k][j];
18 }
19 
20 int main()
21 {
22     cin >> Nv >> Ne;
23     for (int i = 1; i <= Nv; ++i)
24         for (int j = 1; j <= Nv; ++j)
25         {
26             if (i == j)
27                 D[i][j] = 0;
28             else
29                 D[i][j] = INF;
30         }
31 
32     for (int i = 1; i <= Ne; ++i)
33     {
34         int V1, V2;
35         cin >> V1 >> V2;
36         D[V1][V2] = 1;
37     }
38 
39     Floyd();
40 
41     int getRes[MAXVERTEXNUM];
42     memset(getRes, 0, sizeof(getRes));
43     for (int i = 1; i <= Nv; ++i)
44     {
45         for (int j = 1; j <= Nv; ++j)
46         {
47             if (i != j && D[i][j] != INF)
48             {
49                 getRes[j]++;
50                 getRes[i]++;
51             }
52         }
53     }
54 
55     int res = 0;
56     for (int i = 1; i <= Nv; ++i)
57         if (getRes[i] == Nv - 1)
58             res++;
59 
60     cout << res << endl;
61 
62     return 0;
63 }