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
5 5 4 3 4 2 3 2 1 2 2 5
Sample Output
2
题意：有n头奶牛，依次标号为1~n，每一头奶牛都有一个能力值（不重复）。现在两两之间不重复的进行m场battle，自然，能力值高的奶牛将战胜能力值低的奶牛。给出m场battle的最终胜负结果（a，b）（a为胜者），问：根据已知结
果，有多少头奶牛在整体中的排名能被确定。
思路：首先得明确一个点，。最短路一般是求最短的路径或时间或其他，而这个题却是用
最短路的思想来判断两头奶牛之间的关联，首先一个数组ma用于存放念头奶牛之间是否有关联，因为battle有胜负之分，所以关联是单向的。在Floyd算法中只需要判断两点之间是否有单项的关联，最后判断名次时，要明白
一头奶牛在整体中的排名能被确定 <=> 它能被x头不同的奶牛直接或间接打败，同时也能直接或间接打败y头奶牛，也就是它与其他奶牛有双向的关联。
代码：
1 #include <cstdio>
2 #include <fstream>
3 #include <algorithm>
4 #include <cmath>
5 #include <deque>
6 #include <vector>
7 #include <queue>
8 #include <string>
9 #include <cstring>
10 #include <map>
11 #include <stack>
12 #include <set>
13 #include <sstream>
14 #include <iostream>
15 #define mod 998244353
16 #define eps 1e-6
17 #define ll long long
18 #define INF 0x3f3f3f3f
19 using namespace std;
20
21 int main()
22 {
23 //n头奶牛，m场对决
24 int n,m;
25 //ma用于存放两头奶牛之间是否有关系。
26 int ma[110][110];
27 scanf("%d %d",&n,&m);
28 //初始化为0表示都没有关系
29 memset(ma,0,sizeof(ma));
30 int a,b;
31 for(int i=1;i<=m;i++)
32 {
33 scanf("%d %d",&a,&b);
34 //为1表示a和b之间有关系，而且是a到b单向的关系
35 ma[a][b]=1;
36 }
37 //Floyd算法核心
38 for(int k=1;k<=n;k++)
39 {
40 for(int i=1;i<=n;i++)
41 {
42 for(int j=1;j<=n;j++)
43 {
44 //如果i到k有关系，并且k带j有关系，则i可以与j有关系
45 if(ma[i][k]&&ma[k][j])
46 {
47 ma[i][j]=1;
48 }
49 }
50 }
51 }
52 //判断排位名次
53 int ans=0;
54 for(int i=1;i<=n;i++)
55 {
56 int num=1;
57 for(int j=1;j<=n;j++)
58 {
59 //如果i能到j，或j能到i，则两者之间有关系
60 if(ma[i][j]||ma[j][i])
61 {
62 num++;
63 }
64 }
65 //如果i与其他奶牛都有关系，则可以判断i的名次
66 if(num==n)
67 {
68 ans++;
69 }
70 }
71 printf("%d\n",ans);
72 }