Silver Cow Party Time Limit: 2000MS   Memory Limit: 65536K Total Submissions:28457   Accepted: 12928 Description One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X …

Description One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; road i re…

题目链接: https://vjudge.net/problem/POJ-3268 One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads conn…

题目大意: 有N个农场每个农场要有一头牛去参加一个聚会,连接每个农场有m条路, 聚会地点是X,并且路是单向的.要求的是所有牛赶到聚会地点并且回到自己原先的农场所需要的最短时间. 题目分析: 其实就是以X为终点,求出X到其他每个点的距离, 再将图反存一下,在做一次最短路, 两次距离相加求出最长的时间. 这里是用Dijkstra写的,我们第一次用邻接矩阵写,第二次用邻接表,并且有优先队列优化 #include <iostream> #include <cmath> #include &…

POJ 3268 Silver Cow Party Description One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects…

POJ 3268 Silver Cow Party (最短路径) Description One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads c…

题目链接:http://poj.org/problem?id=3268 Silver Cow Party Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 19211   Accepted: 8765 Description One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow…

原题链接:http://poj.org/problem?id=3268 Silver Cow Party Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 15545   Accepted: 7053 Description One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow…

Silver Cow Party Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 3268 Description One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to b…

题目传送门 1 2 题意:有向图,所有点先走到x点,在从x点返回,问其中最大的某点最短路程 分析:对图正反都跑一次最短路,开两个数组记录x到其余点的距离,这样就能求出来的最短路以及回去的最短路. POJ 3268 //#include <bits/stdc++.h> #include <cstdio> #include <queue> #include <algorithm> #include <cstring> using namespace…

Silver Cow Party Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 13982   Accepted: 6307 Description One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X …

Silver Cow Party 题目链接: http://acm.hust.edu.cn/vjudge/contest/122685#problem/D Description One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤…

Silver cow party 迪杰斯特拉+反向 题意 有n个农场,编号1到n,每个农场都有一头牛.他们想要举行一个party,其他牛到要一个定好的农场中去.每个农场之间有路相连,但是这个路是单向的,并且去了还得回来,求花费时间最多是多少? 解题思路 很容易想明白需要分两步 第一步:算出目的点到其他点的最短距离, 这一步很好实现,直接使用Dijkstra即可 第二部:算出其他点到终点的最短距离. 关键就在第二部.迪杰斯特拉算的是某一点到其他所有点的最短距离,而这次我们是求的其他点到某一点的最短…

                                                                                                   Silver Cow Party Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 19325   Accepted: 8825 Description One cow from each of N farms (1 ≤ N ≤…

Silver Cow Party Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other) Total Submission(s) : 1   Accepted Submission(s) : 1 Problem Description One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is goin…

Silver Cow Party Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 13611   Accepted: 6138 Description One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X…

Silver Cow Party One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; road…

Silver Cow Party Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 22864   Accepted: 10449 Description One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X…

Silver Cow Party Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 12674   Accepted: 5651 Description One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X …

Silver Cow Party Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 17017   Accepted: 7767 Description One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X …

描述: One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; road i requires T…

题意:有n个地点,有m条路,问从所有点走到指定点x再走回去的最短路中的最长路径 思路:用Floyd超时的,这里用的Dijkstra. Dijkstra感觉和Prim和Kruskal的思路很像啊.我们把所有点分为两个集合:S(和源点在同一集合),T(其余点),用dis数组表示每个点到S的最短距离,vis数组记录这个点是否在S中.我们每次找出在T的一个和S距离最短的点,加到S中,这个距离就是他到源点的最短路径(听说能证,我不会证...),然后更新其他点,直到所有点都在S. 这道题从x走回去很简单,就…

题意:由n个牧场,编号1到n.每个牧场有一头牛.现在在牧场x举办party,每头牛都去参加,然后再回到自己的牧场.牧场之间会有一些单向的路.每头牛都会让自己往返的路程最短.问所有牛当中最长的往返路程是多少. 思路:n最多到1000,floyd肯定超时.可以这样做,把图中所有的边先存起来,然后第一次用dijkstra求出以x为源点到每个点的最短距离.该最短距离为每头牛回家时的最短距离.然后建个新的图,将之前存的边反向加入图中.如之前有条从5到8距离为2的路,则此时向图中添加的边为从8到5距离为2的…

本题链接 : http://poj.org/problem?id=3268 题目大意:牛们要去聚会,输入N = 顶点数(牛场):M = 边(路)的数目: X = 终点 (聚会点).问题:求来回时间的最大值. Description: One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1…

http://poj.org/problem?id=3268 每头牛都要去标号为X的农场参加一个party,农场总共有N个(标号为1-n),总共有M单向路联通,每头牛参加完party之后需要返回自己的农场,但是他们都想选一条最近的路,并且由于路是单向的,去的路和来的路选择可能不一样,问来去时间之和最大是多少? 这题等于给定了起点和终点,那么求出(d[x][i]+d[i][x])最大的那个即可. 开始错了几次,太不小心了,就是最后求最大值的时候,用了一个临时变量没置0,所以可能会导致错误. #in…

题目 Dijkstra,正反两次最短路,求两次和最大的. #define _CRT_SECURE_NO_WARNINGS //这是找出最短路加最短路中最长的来回程 //也就是正反两次最短路相加找最大的和 #include<string.h> #include<stdio.h> #include<math.h> #include<algorithm> using namespace std; ; #define typec int const typec IN…

题目:click here 题意: 给出n个点和m条边,接着是m条边,代表从牛a到牛b需要花费c时间,现在所有牛要到牛x那里去参加聚会,并且所有牛参加聚会后还要回来,给你牛x,除了牛x之外的牛,他们都有一个参加聚会并且回来的最短时间,从这些最短时间里找出一个最大值输出. 分析: 最短路径只需要从x到i的最短路径代表他们返回的最短路径,然后将所有边反过来,再从x到i的最短路径代表他们来参加聚会的最短路径,这样对应相加找出一个最大值就可以了,当然其实不需要将所有边反过来,只需要将map的行和列对换一…

题意: 在一个有向图中求n头牛从自己的起点走到x再从x走回来的最远距离 思路一开始是暴力跑dij…… 讲道理不太可能…… 然后就百度了一下 才知道把矩阵转置的话就只需要求两次x的单源最短路…… /* *********************************************** Author :Sun Yuefeng Created Time :2016/10/22 20:09:36 File Name :A.cpp *******************************…

(￣▽￣)" #include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #include<queue> using namespace std; const int INF=10e8; ; int k,minn; int cost[MAXN][MAXN],lowcost[MAXN],lc[MAXN]; bool vis[MAXN]; void Re…

题目:http://poj.org/problem?id=3268 题解:使用 priority_queue队列对dijkstra算法进行优化 #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <functional> #include <cstring> using namespace std; + ; ; str…