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…
Six Degrees of Cowvin Bacon Descriptions 数学课上,WNJXYK忽然发现人缘也是可以被量化的,我们用一个人到其他所有人的平均距离来量化计算. 在这里定义人与人的距离:1.自己与自己的距离为02.如果A和B属于同一个小团体,那么他们之间的距离为13.如果A与B属于一个小团体,B与C属于一个小团体,且A与C不同属于任何一个小团体,那么A与C的距离为2(A联系C,经过B.C两个人)4.以此类推 班里有N个人 (2 <= N <= 300),共有M对小团体关系(…
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…
Description An ascending sorted sequence of distinct values is one in which some form of a less-than operator is used to order the elements from smallest to largest. For example, the sorted sequence A, B, C, D implies that A < B, B < C and C < D.…
题目: BIT has recently taken delivery of their new supercomputer, a 32 processor Apollo Odyssey distributed shared memory machine with a hierarchical communication subsystem. Valentine McKee's research advisor, Jack Swigert, has asked her to benchmark…
Wormholes 题目描述 教学楼里有很多教室,这些教室由双向走廊连接.另外,还存在一些单向的秘密通道,通过它们可以回到过去.现在有 N (1 ≤ N ≤ 500) 个教室,编号 1..N, M (1 ≤ M ≤ 2500) 条走廊,和 W (1 ≤ W ≤ 200) 条秘密通道. DY在养猫之余,还是一个时间旅行爱好者.她希望从一间教室出发,经过一些走廊和秘密通道,回到她出发之前的某个时间. 共有F (1 ≤ F ≤ 5) 组数据.对每组数据,判断DY是否有回到过去的可能性.不存在耗时超过1…
本题链接:http://poj.org/problem?id=2139 Description: The cows have been making movies lately, so they are ready to play a variant of the famous game "Six Degrees of Kevin Bacon". The game works like this: each cow is considered to be zero d…
floyd算法是被大家熟知的最短路算法之一,利用动态规划的思想,f[i][j]记录i到j之间的最短距离,时间复杂度为O(n^3),虽然时间复杂度较高,但是由于可以处理其他相似的问题,有着广泛的应用,这些变形的问题也是考察重点之一. 伪代码大致如下: a) 初始化:D[u,v]=A[u,v] b) For k:=1 to n For i:=1 to n For j:=1 to n If D[i,j]>D[i,k]+D[k,j] Then D[i,j]:=D[i,k]+D[k,j]; c) 算法结束…
1 /* 题目大意: 从一个点到达另一个点有多条路径,求这多条路经中最大噪音值的最小值! . 思路:最多有100个点,然后又是多次查询,想都不用想,Floyd算法走起! */ #include<iostream> #include<cstring> #include<cstdio> #define INF 0x3f3f3f3f using namespace std; ][]; int main(){ int n, m, q; int u, v, d; ; scanf(…