Description FJ的N(1 <= N <= 100)头奶牛们最近参加了场程序设计竞赛:).在赛场上,奶牛们按1..N依次编号.每头奶牛的编程能力不尽相同,并且没有哪两头奶牛的水平不相上下,也就是说,奶牛们的编程能力有明确的排名. 整个比赛被分成了若干轮,每一轮是两头指定编号的奶牛的对决.如果编号为A的奶牛的编程能力强于编号为B的奶牛(1 <= A <= N; 1 <= B <= N; A != B) ,那么她们的对决中,编号为A的奶牛总是能胜出. FJ想知道奶
Description FJ的N(1 <= N <= 100)头奶牛们最近参加了场程序设计竞赛:).在赛场上,奶牛们按1..N依次编号.每头奶牛的编程能力不尽相同,并且没有哪两头奶牛的水平不相上下,也就是说,奶牛们的编程能力有明确的排名. 整个比赛被分成了若干轮,每一轮是两头指定编号的奶牛的对决.如果编号为A的奶牛的编程能力强于编号为B的奶牛(1 <= A <= N; 1 <= B <= N; A != B) ,那么她们的对决中,编号为A的奶牛总是能胜出. FJ想知道奶
floyd传递关系,一个牛能确定排名的条件是能和所有牛确定关系 #include<iostream> #include<cstdio> using namespace std; const int N=105; int n,m,a[N][N],ans; int main() { scanf("%d%d",&n,&m); for(int i=1;i<=m;i++) { int x,y; scanf("%d%d",&
1648: [Usaco2006 Dec]Cow Picnic 奶牛野餐 Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 554 Solved: 346[Submit][Status][Discuss] Description The cows are having a picnic! Each of Farmer John's K (1 <= K <= 100) cows is grazing in one of N (1 <= N <
P2419 [USACO08JAN]牛大赛Cow Contest 题目背景 [Usaco2008 Jan] 题目描述 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 t
直接floyd.. ---------------------------------------------------------------------------- #include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #define rep( i , n ) for( int i = 0 ; i < n ; ++i ) #define clr(
POJ 3660 Cow Contest / HUST 1037 Cow Contest / HRBUST 1018 Cow Contest(图论,传递闭包) 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
Cow Contest Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 16941 Accepted: 9447 题目链接:http://poj.org/problem?id=3660 Description: N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all k