Problem Description

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

題意：給出兩頭牛之間的競爭的結果(a,b)，a贏b，確定有多少頭牛能夠確定其排名。

思路：對于一頭牛，如果想確定其排名，那么就是在明確的知道比它高的有幾只、低的有幾只，且前后加起來正好是 n-1 只時，可以確定他的排名。

所給的關系可以看做是一條路，只需判斷任意兩點是否連通，再確定一個點的入度和出度相加是否等于 n-1，因此可以判斷這個點是否為所求的點。

因此本題實質是判斷一個有向無權圖的任意兩點是否導通，即求圖的傳遞閉包，使用 Floyd 即可。

Source Program

#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<string> #include<cstdlib> #include<queue> #include<vector> #define INF 0x3f3f3f3f #define PI acos(-1.0) #define N 101 #define MOD 2520 #define E 1e-12 using namespace std; int g[N][N]; int degree[N][N]; int main() {int n,m;while(scanf("%d%d",&n,&m)!=EOF){memset(g,0,sizeof(g));for(int i=0;i<m;i++){int x,y;scanf("%d%d",&x,&y);g[x][y]=1;}for(int k=1;k<=n;k++)for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)if(g[i][k]&&g[k][j])g[i][j]=1;memset(degree,0,sizeof(degree));for(int i=1;i<=n;i++)//計算點的度數{for(int j=1;j<=n;j++)if(i!=j){if(mp[i][j]){degree[i]++;degree[j]++;}}}int cnt=0;for(int i=1;i<=n;i++)if(degree[i]==n-1)cnt++;printf("%d\n",cnt);}return 0; }

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