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

?

#include<stdio.h> #include <string.h> int g[110][120],n,m,i,j,k; void ff() {for(k=1; k<=n; k++)for(i=1; i<=n; i++)for(j=1; j<=n; j++)g[i][j]=g[i][j] || (g[i][k]&&g[k][j]); } int main() {int u, v;memset(g, 0, sizeof(g));scanf("%d%d", &n,&m);while(m--){scanf("%d%d", &u,&v);g[u][v] = 1;}ff();int s, e=0;for(i=1; i<=n; i++){s=0;for(j=1; j<=n; j++){if(g[i][j] || g[j][i])s++;}if(s==n-1)e++;}printf("%d", e);return 0; }

題意? ?一群牛比賽，每場(chǎng)兩只牛對(duì)打，并分出勝負(fù)，現(xiàn)在問你能確定幾只牛的戰(zhàn)斗力排名。

思 利用Folyed 來找找出每頭牛可以和多少頭牛建立關(guān)系，當(dāng)一頭牛可以和剩下的N-1頭牛都可以建立關(guān)系時(shí)，它的名次就可以 確定了

總結(jié)

以上是生活随笔為你收集整理的ayit第十六周周赛题 a题的全部?jī)?nèi)容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網(wǎng)站內(nèi)容還不錯(cuò)，歡迎將生活随笔推薦給好友。