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…

题意:输出所有的环: 思路:数据比较小,用三层循环的floyd传递闭包(即两条路通为1,不通为0,如果在一个环中,环中的所有点能互相连通),输出路径用dfs,递归还没有出现过的点(vis),输出并递归该点与其他点能互达的点: #include <cstdio> #include <vector> #include <string> #include <cstring> #include <iostream> using namespace std…

UNIX 插头 紫书P374 [题目链接]UNIX 插头 [题目类型]EK网络流+Floyd传递闭包 &题解: 看了书之后有那么一点懂了,但当看了刘汝佳代码后就完全明白了,感觉他代码写的好牛逼啊,Orz 所以就完全照着码了一份. [时间复杂度]O(\(n^3\)) &代码: #include <iostream> #include <cstring> #include <string> #include <vector> #include &…

电话圈 紫书P365 [题目链接]电话圈 [题目类型]Floyd传递闭包+输出连通分量 &题解: 原来floyd还可以这么用,再配合连通分量,简直牛逼. 我发现其实求联通分量也不难,就是for循环+dfs+vis记录数组. 在发上刘汝佳的代码链接:https://github.com/aoapc-book/aoapc-bac2nd/blob/master/ch11/UVa247.cpp [时间复杂度]O(\(n^3\)) &代码: #include <bits/stdc++.h>…

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…

Treasure Exploration Time Limit:6000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 2594 Description Have you ever read any book about treasure exploration? Have you ever see any film about treasure exploratio…

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…

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…

Treasure Exploration Time Limit: 6000MS   Memory Limit: 65536K Total Submissions: 8130   Accepted: 3325 Description Have you ever read any book about treasure exploration? Have you ever see any film about treasure exploration? Have you ever explored…