title: Cow Contest 弗洛伊德+传递闭包 nyoj211 tags: [弗洛伊德,传递闭包] 题目链接 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 rati…

Cow Contest 时间限制:1000 ms  |  内存限制:65535 KB 难度:4   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 that is…

Cow Contest DescriptionN (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.…

题目链接:http://poj.org/problem?id=3660 Cow Contest Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13085   Accepted: 7289 Description N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all kn…

题目链接:https://cn.vjudge.net/problem/POJ-3660 题意 有n头牛,每头牛都有一定的能力值,能力值高的牛一定可以打败能力值低的牛 现给出几头牛的能力值相对高低 问在一场一对一的比赛中,那些牛的排名可以确定下来 思路 一开始还以为是topo排序,每次去掉没有入度或出度的节点 若有两个及以上的节点可以去掉,则排序结束 然后写出来WA两发... 正确思路: 若满足x头牛可以打败牛a,牛a可以打败y头牛,且n==x+y-1时牛a排名唯一确定 那么可以利用Floyd传递…

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 contes…

题目链接:http://acm.nyist.net/JudgeOnline/problem.php?pid=211 思路:我的思路是对每一个点,向上广搜,向下广搜,看总共能不能搜到n-1个结点,能,表明该节点的等级确定.  题目标程给的方法是floyd+统计,似乎简单的多,我弄复杂了,还是贴个代码吧! 代码如下: #include "stdio.h" #include "string.h" #include "queue" using names…

链接 Cow Contest Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu 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. Eac…

Cow Contest Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5989   Accepted: 3234 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…

解题思路:给出n头牛,和这n头牛之间的m场比赛结果,问最后能知道多少头牛的排名. 首先考虑排名怎么想,如果知道一头牛打败了a头牛,以及b头牛打赢了这头牛,那么当且仅当a+b+1=n时可以知道排名,即为此时该牛排第b+1名. 即推出当一个点的出度和入度的和等于n-1的时候,该点的排名是可以确定的, 即用传递闭包来求两点的连通性,如果d[i][j]==1,那么表示i,j两点相连通,度数都分别加1 Cow Contest Time Limit: 1000MS   Memory Limit: 65536…

/* floyd 传递闭包 开始Floyd 之后统计每个点能到的或能到这个点的 也就是他能和几个人确定胜负关系 第一批要有n-1个 然后每次减掉上一批的人数 麻烦的很 复杂度上天了.... 正难则反 我们考虑一定不能确定排名的 */ #include<iostream> #include<cstdio> #include<cstring> #define maxn 110 using namespace std; int n,m,f[maxn][maxn],ans; i…

链接:poj 3660 题意:给定n头牛,以及某些牛之间的强弱关系.按强弱排序.求能确定名次的牛的数量 思路:对于某头牛,若比它强和比它弱的牛的数量为 n-1,则他的名次能够确定 #include<stdio.h> #include<limits.h> int a[110][110]; int main() { int n,m,i,j,k,s,sum; while(scanf("%d%d",&n,&m)!=EOF){ for(i=1;i<=…

题意: n个点,m条边. 若A 到 B的边存在,则证明 A 的排名一定在 B 前. 最后求所有点中,排名可以确定的点的个数. n <= 100, m <= 4500 刚开始还在想是不是拓扑排序. n这么小的数据范围,典型的传递闭包.直接可以用Floyd求. 求出传递闭包之后找哪些点与其他所有点都有直接关系,即所有的点和它之间排名都可以确定.计入答案. #include <iostream> #include <cstdlib> #include <cstdio&g…

题目链接:http://poj.org/problem?id=3660 题意是给你n头牛,给你m条关系,每条关系是a牛比b牛厉害,问可以确定多少头牛的排名. 要是a比b厉害,a到b上就建一条有向边......这样建好之后,如果比a牛厉害的牛都能达到a牛,而a牛能到达比a牛差的牛的话,牛的头数又恰好是n-1头,那么a牛的排名则是确定的. 所以用Flody比较方便,要是i到k能到达,k到j能到达,那么i到j就能到达,i就比j厉害. #include <iostream> #include <…

P2419 [USACO08JAN]牛大赛Cow Contest Floyd不仅可以算最短路,还可以处理点之间的关系. 跑一遍Floyd,处理出每个点之间是否有直接或间接的关系. 如果某个点和其他$n-1$个点都有关系,那么它的排名就是可确定的. #include<iostream> #include<cstdio> #include<cstring> #define re register using namespace std; ][],ans; int main(…

Cow Contest Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 16341   Accepted: 9146 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 other…

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…

Cow Contest Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 3660 Description N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some co…

原题链接:http://poj.org/problem?id=3660 Cow Contest Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8395   Accepted: 4734 Description N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all kno…

题目链接:http://www.lydsy.com/JudgeOnline/problem.php?id=1612 题意: 有n头牛比赛. 告诉你m组(a,b),表示牛a成绩比牛b高. 保证排名没有并列. 问你有多少只牛的排名已经确定. 题解: 对于一头牛,它的排名确定的条件是:它前面的牛数量 + 它后面的牛数量 = n-1 所以对于(a,b),连一条有向边a->b. 然后做floyd传递闭包. 枚举每一头牛,统计与它连通的牛的个数sum. 如果sum = n-1,则ans++. AC Code…

1612: [Usaco2008 Jan]Cow Contest奶牛的比赛 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 891  Solved: 590[Submit][Status][Discuss] 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…

Cow Contest 题目链接: http://acm.hust.edu.cn/vjudge/contest/122685#problem/H 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 certa…

对于第 i 头牛 , 假如排名比它高和低的数位 n - 1 , 那么他的 rank 便可以确定 . floyd --------------------------------------------------------------------------------------- #include<cstdio> #include<cstring> #include<algorithm> #include<iostream>   #define rep…

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: 10450   Accepted: 5841 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 other…

Cow Contest POJ - 3660 :http://poj.org/problem?id=3660   参考:https://www.cnblogs.com/kuangbin/p/3140837.html   题意: n头牛,有m对牛进行了比赛,现在告诉你每队牛比赛的结果,A胜B,问有几头牛的排名可以确定. 思路: 题目给出了m对的相对关系,求有多少个排名是确定的. 使用floyed求一下传递闭包.如果这个点和其余的关系都是确定的,那么这个点的排名就是确定的. #include <al…

  Cow Contest Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 17797   Accepted: 9893 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 oth…

题意:给出m个关系,问你能确定机头牛的排名 思路:要确定排名那必须要把他和其他n-1头牛比过才行,所以Floyd传递闭包,如果赢的+输的有n-1就能确定排名. 代码: #include<cstdio> #include<set> #include<map> #include<cmath> #include<stack> #include<vector> #include<queue> #include<cstring…

Cow Contest 时间限制:1000 ms  |  内存限制:65535 KB 难度:4 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 that is u…

题目链接:http://poj.org/problem?id=3660 Cow Contest Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10066   Accepted: 5682 Description N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all kn…