POJ 3660 Cow Contest

Cow Contest

Time Limit: 1000ms

Memory Limit: 65536KB

This problem will be judged on PKU. Original ID: 3660
64-bit integer IO format: %lld Java class name: Main

N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some cows code better than others. Each cow has a certain constant skill rating that is unique among the competitors.

The contest is conducted in several head-to-head rounds, each between two cows. If cow A has a greater skill level than cow B (1 ≤ A ≤ N; 1 ≤ B ≤ N; A ≠ B), then cow A will always beat cow B.

Farmer John is trying to rank the cows by skill level. Given a list the results of M (1 ≤ M ≤ 4,500) two-cow rounds, determine the number of cows whose ranks can be precisely determined from the results. It is guaranteed that the results of the rounds will not be contradictory.

Input

* Line 1: Two space-separated integers: N and M
* Lines 2..M+1: Each line contains two space-separated integers that describe the competitors and results (the first integer, A, is the winner) of a single round of competition: A and B

Output

* Line 1: A single integer representing the number of cows whose ranks can be determined　

Sample Input

Sample Output

Source

USACO 2008 January Silver

解题：传递闭包，一头牛的排名是确定的当且仅当比他弱的牛的头数跟比他强的牛的头数的和为n-1

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <climits>
 7 #include <vector>
 8 #include <queue>
 9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #include <stack>
13 #define LL long long
14 #define pii pair<int,int>
15 #define INF 0x3f3f3f3f
16 using namespace std;
17 const int maxn = 101;
18 bool g[maxn][maxn];
19 int n,m;
20 void Floyd() {
21     for(int k = 1; k <= n; ++k) {
22         for(int i = 1; i <= n; ++i) {
23             if(g[i][k]) {
24                 for(int j = 1; j <= n; ++j)
25                     if(!g[i][j]) g[i][j] = g[i][k]&&g[k][j];
26             }
27         }
28     }
29 }
30 int main() {
31     while(~scanf("%d %d",&n,&m)) {
32         memset(g,false,sizeof(g));
33         for(int i = 0; i < m; ++i) {
34             int u,v;
35             scanf("%d %d",&u,&v);
36             g[u][v] = true;
37         }
38         Floyd();
39         int ans = 0,x,y;
40         for(int i = 1; i <= n; ++i){
41             x = y = 0;
42             for(int j = 1; j <= n; ++j){
43                 if(g[i][j]) x++;
44                 if(g[j][i]) y++;
45             }
46             if(x+y == n-1) ans++;
47         }
48         printf("%d
",ans);
49     }
50     return 0;
51 }

View Code