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

有n只奶牛，有n個(gè)連續(xù)的實(shí)力，如果u的實(shí)力大于v的實(shí)力，就能打贏它，
然后給定m種關(guān)系，求最后能確定其排名的奶牛個(gè)數(shù)。

一個(gè)傳遞閉包問題。頭一次解這種題目。謝謝大神的思路。

//Asimple #include <iostream> #include <algorithm> #define mod 100000 #define CLS(a, v) memset(a, v, sizeof(a)) #define debug(a) cout << #a << " = " << a <<endl #define dobug(a, b) cout << #a << " = " << a << " " << #b << " = " << b << endl using namespace std; typedef long long ll; const int maxn = 500+5;const int INF = (1 << 16);int n, m, num, T, k, len, ans, sum, x, y, z; int Map[maxn][maxn]; void solve(){for(int k=1; k<=n; k++)for(int i=1; i<=n; i++)for(int j=1; j<=n; j++)if( Map[i][k] && Map[k][j] )Map[i][j] = 1;/*傳遞閉包 只有這個(gè)點(diǎn)和其余所有的點(diǎn)的關(guān)系都是確定的這個(gè)點(diǎn)才是確定的*/ans = 0;int j;for(int i=1; i<=n; i++) {for(j=1; j<=n; j++) {if( i==j ) continue;if( Map[i][j]==0 && Map[j][i]==0) break;}if( j>n ) ans ++;}cout << ans << endl; }void input() {ios_base::sync_with_stdio(false);while( cin >> n >> k ) {CLS(Map, 0);while( k -- ) {cin >> x >> y;Map[x][y] = 1;}solve();} }int main(){input();return 0; }

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轉(zhuǎn)載于:https://www.cnblogs.com/Asimple/p/6915788.html

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