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…

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…

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…

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…

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 /* Name:nyoj-211-Cow Contest Copyright: Author: Date: 2018/4/27 21:02:06 Description: floyd算法 大佬的惊奇思路 */ #include <iostream> #include <cstdio> #include <cstring> using namespace std; ; const int INF = 0x3f3f3f3f; int N, g[MAXN][MAXN…

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…

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: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…

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(…

题目链接:http://poj.org/problem?id=3660 Description N ( ≤ N ≤ ) cows, conveniently numbered ..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 am…

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…

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…

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: 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…

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…

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…

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…

Floyd 算法小结  By Wine93 2013.11 1. Floyd算法简介 Floyd算法利用动态规划思想可以求出任意2点间的最短路径,时间复杂度为O(n^3),对于稠密图, 效率要高于执行|V|次Dijkstra算法. 核心代码如下: for(k=1;k<=n;k++) for(i=1;i<=n;i++) for(j=1;j<=n;j++) dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]); 相关应用 : 有向图:①求任意2点间最短路径…

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…

POJ3660 Cow Contest 题目链接:http://poj.org/problem?id=3660 题意:农名约翰有些奶牛,约翰通过让他们决斗来决定他们的排名,约翰让这些奶牛一对一打完一定的局数之后,问有哪些奶牛的排名是可以确定的(注:a打得过b,b打得过c,则a打得c) 根据题意我们明白当一个奶牛和其他的所有奶牛都存在胜负关系的时候,他的排名就是确定的. 思路:利用floyd算法,传递闭包,算出可达矩阵.然后在每一行去查询该点和其他所有的点的联系是不是n-1. 代码: //Auth…

Floyd 判断连通性 d[i][j]仅表示i,j之间是否联通 ;k<=n;k++) ;i<=n;i++) ;j<=n;j++) dis[i][j]=dis[i][j]||(dis[i][k]&&dis[k][j]); 有向图和无向图都适用 当然了,也可以DFS判断连通性 裸题: P2419 [USACO08JAN]牛大赛Cow Contest 题目背景 [Usaco2008 Jan] 题目描述 N (1 ≤ N ≤ 100) cows, conveniently num…

目录 一.Floyd算法 二.Floyd算法的应用 1. 传递闭包 例1:P2881 [USACO07MAR]排名的牛Ranking the Cows 例2:P2419 [USACO08JAN]牛大赛Cow Contest 2.快速求出多源最短路 例1:P1522 牛的旅行 Cow Tours 3.解决双权值问题 例1:P1119 灾后重建 一.Floyd算法 如何求任意两点最短路?我们可以运行n次SPFA或Dijkstra求得, 而Floyd算法能在\(O(N^3)\)的时间复杂度内求出图中任…

Cow Contest POJ-3660 1.本题考察的是最短路,用的算法是Floyd算法 2.如果一个结点和剩余的n-1个结点都有关系,那么可以确定其排名 3.需要注意的是,判断是否有关系时,反向关系也要考虑 #include<iostream> #include<cstdio> #include<algorithm> #include<cstring> using namespace std; int n,m; int map[101][101]; in…

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…

Cow Contest Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7690   Accepted: 4288 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…

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…