Silver Cow Party Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 15156   Accepted: 6843 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 http://poj.org/problem?id=3268 题意 有向图中有n个结点,编号1~n,输入终点编号x,求其他结点到x结点来回最短路长度的最大值. 思路 最短路问题,有1000个结点,Floyd算法应该会超时,我刚开始使用的Dijkstra算法也超时,原因是因为我使用一个循环遍历结点1~n,每次遍历我都使用两次Dijkstra求i到x和x到i的最短路,时间复杂度太高.降低时间复杂度的方法是先在原矩阵的基础上使用dijkstra求结点x到其余各点的最短路径,然后将矩阵转置,在…

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 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…

题目链接:http://poj.org/problem?id=3268 Silver Cow Party Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 24527   Accepted: 11164 Description One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big co…

题目链接: https://vjudge.net/problem/POJ-3268 题目大意: 有编号为1-N的牛,它们之间存在一些单向的路径.给定一头牛的编号X,其他牛要去拜访它并且拜访完之后要返回自己原来的位置,求所有牛从开始到回家的时间是多少? 思路: 所有牛都回到了家所花费的时间就是这些牛中花费时间的最大者,可以正向的Dijkstra求出从X到每个点的最短时间,然后一个骚操作:将所有边反向(swap(Map[i][j], Map[j][i])),再从x用dijkstra求出X到每个点的最…

https://vjudge.net/problem/POJ-3268 一开始floyd超时了.. 对正图定点求最短,对逆图定点求最短,得到任意点到定点的往返最短路. #include<iostream> #include<cstdio> #include<queue> #include<cstring> #include<algorithm> #include<cmath> #include<stack> #define…

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 Ti (1…

题目描述 原题来自:USACO 2007 Feb. Silver N(1≤N≤1000)N (1 \le N \le 1000)N(1≤N≤1000) 头牛要去参加一场在编号为 x(1≤x≤N)x(1 \le x \le N)x(1≤x≤N) 的牛的农场举行的派对.有 M(1≤M≤100000)M(1\le M \le 100000)M(1≤M≤100000) 条有向道路,每条路长 Ti(1≤Ti≤100)T_i(1 \le T_i \le 100)T​i​​(1≤T​i​​≤100):每头牛都…

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 Ti (1…

有n个农场,编号1~N,农场里奶牛将去X号农场.这N个农场之间有M条单向路(注意),通过第i条路将需要花费Ti单位时间.选择最短时间的最优路径来回一趟,花费在去的路上和返回农场的这些最优路径的最长时间是多少? 思路:计算出每头牛去X并且回来的最短路径所需要的时间,然后求出这n-1个农场的牛的最长时间即可,两次运用dijkstra: 1 计算回来的时间:以X为源点,求出源点到各个农场的最短路径: 2 计算去的时间:将路径反转,在用一次djk,求源点到农场的最短路径(实则求牛去X的最短路径): 3…

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…

http://poj.org/problem?id=3268 题目大意:求到x距离与从x返回和的最大值 从x点到各个点最短路好求,直接用Dijkstar,但从各个点到x点却不好求,只要把路向翻转过来也变成求从x点到各个点,直接用Dijstar dist[]记录x点到各个点的最短路径距离 #include<stdio.h> #include<stdlib.h> #include<math.h> #include<string.h> #define max(a,…

其实还是从一个x点出发到所有点的最短路问题.来和回只需分别处理一下逆图和原图,两次SPFA就行了. #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<string> #include<cmath> #include<map> #include<set> #include<vector> #…

非常感谢kuangbin专题啊,这道题一开始模拟邻接表做的,反向边不好处理,邻接矩阵的话舒服多了. 题意:给n头牛和m条有向边,每头牛1~n编号,求所有牛中到x编号去的最短路+回来的最短路的最大值. #include <iostream> #include <cstring> #include <cstdio> #include <algorithm> #include <queue> using namespace std; ; ; const…

Silver Cow Party Descriptions 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 Input 第1行:三个空格分隔的整数,分别为: N, M和 X 行2 .. M +1:行 i +1描述具有三个空格分隔整数的道路 i: A i, B i和 T i.所描述的道路从农场A i运行 到农场 B…

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…

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…

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 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 …

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…

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…

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…

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…

Silver Cow Party POJ-3268 这题也是最短路的模板题,只不过需要进行两次求解最短路,因为涉及到来回的最短路之和. 该题的求解关键是:求解B-A的最短路时,可以看做A是起点,这就和求解A-B的最短路很类似了,只不过需要将单向路的距离调换一下即可. #include<iostream> #include<cstdio> #include<algorithm> using namespace std; const int INF = 1111111111…

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

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 *******************************…

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…

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…