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…

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…

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…

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…

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…

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…

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<=…

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…

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…

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…

Cow Contest Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11129   Accepted: 6183 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…

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…

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…

http://poj.org/problem?id=3660 题目大意:n头牛两两比赛经过m场比赛后能判断名次的有几头可转 化为路径问题,用Floyd将能够到达的路径标记为1,如果一个点能 够到达剩余你n - 1个点则能判断该点的名次 #include<stdio.h> #include<stdlib.h> #include<math.h> #include<string.h> #include<queue> #define INF 0x3f3f…

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 <…

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…

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 conduct…

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 conduct…

题目链接: http://poj.org/problem?id=3660 题目大意: 给出n头牛,m个关系,关系为a的战力比b高.求最后可以确定排名的牛的数量 思路: 1.如果一头牛跟其他所有牛都确定了一个输赢关系,那么该牛的排名就得到了确定,所以用floyd跑一遍传递闭包.然后求得每个点的出度(赢了), 入度(输了).若该点的度之和为 n - 1 ,即确定排名. #include<stdio.h> #include<string.h> #define mem(a, b) mems…

传送门:http://poj.org/problem?id=3660 题意:有n头牛, 给你m对关系.(a, b)表示牛a能打败牛b, 求在给出的这些关系下, 能确定多少头牛的排名. 传递闭包: 关系之间具有传递性(例如a> b, b> c, 那么a> c), 在那些已给出的关系基础上, 通过传递性, 把所有可能的关系都找出来. 思路:假设一头牛可以被X头牛打败,可以打败Y头牛.如果这头牛的排名可以确定则X+Y=N-1.想到Floyd正好可以求X→Y的关系,只是将Floyd求距离改成了判…

MPI Maelstrom 题目链接: http://acm.hust.edu.cn/vjudge/contest/122685#problem/G Description BIT has recently taken delivery of their new supercomputer, a 32 processor Apollo Odyssey distributed shared memory machine with a hierarchical communication subsy…

题意难懂是POJ的标配,这都TM赖本泽马. 题意:有N头牛进行了M场比赛,比赛双方是A - B 且总是A赢(前面的那个数赢),如果说A赢B,B赢C 则可以确定A赢C.问最终多少头牛的排名可以确定. 思路:用弗洛伊德判读每两点之间是否有输赢关系(INF表示没关系),如果某个点与所有点都有输赢关系,则这个点的排名可以确定. #include<stdio.h> #include<string.h> #include<cstring> #include<string>…

<题目链接> 题目大意: 有n头牛, 给你m对关系(a, b)表示牛a能打败牛b, 求在给出的这些关系下, 能确定多少牛的排名. 解题分析: 首先,做这道题要明确,什么叫确定牛的排名.假设该牛被x头牛打败(直接或间接),同时它也有y头手下败将(直接或间接),当x+y=n-1时,即除这头牛本身外,其他所有的牛都为这头牛贡献了出度或者入度.即,当这头牛与其它所有的牛的输赢关系都确定时(直接或间接),这头牛的排名也就可以确定了.而题目只给出了一些牛的直接输赢关系,这时,我们就可以利用Floyed算法…

题目传送门 题意: m组关系,A能打败B,问最后有几头牛的排名能确定 分析:如果排名确定,那么能打败它的到它一定通,它到能打败的一定能通,也就是和为n-1.用Floyd的传递闭包 #include <cstdio> #include <algorithm> #include <cstring> using namespace std; const int N = 1e2 + 5; bool d[N][N]; int in[N]; int n, m; void Floyd…

/* 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…

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…

题意: 给定n头牛, 然后有m个比较, 求出有多少头牛能确定自己的排名. 分析: 假设有一头牛a, 有ki头牛强于自己, kj头牛弱于自己, ki + kj == n-1时, 那么这头牛的排名就确定了. 对于每个比较建一条有向边 求出a点可达哪些点, 哪些点可达a点即可 #include<cstdio> #include<string> #include<iostream> #include<cstring> #include<vector> #…

题意:牛之间有绝对的强弱,给出一些胜负关系,问有多少头牛可以确定其绝对排名. #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <vector> #include <queue> #include <stack> #include <map> #i…