思路:有向图.假设在X牧场参加party,从X回家的时候,以X为起点,使用一次Dijkstra算法即可.难点在于去X参加party的最短路如何求解. 这时候我们可以反向建图,即把原来有向图的方向全部反向,形成一幅新的有向图G',此时再对G'使用一次以X为起点的Dijkstra算法即    可求得原图G中其他各点以X为终点的最短路径. #include<iostream> #include<vector> #include<string> #include<cmat…

题目链接:https://vjudge.net/problem/POJ-3268 题意:点X处开办排队,其他点的牛到X点去参加派对,然后从X点回到各自的点,通路是单向的,所有牛都要走最短路, 求出所有牛走的最短路中最长的那条路. (直接看代码吧,就是一处改一下,别的和dijkstra板子差不多) #include <iostream> #include <cstring> #include <algorithm> #include <cstdio> #inc…

牛的聚会 题目大意:一群牛在一块农田的不同的点,现在他们都要去到同一个地方开会,然后现在从那个地方回到原来的位置,点与点之间的连线都是单向的,并且通过一个路径需要一定时间,问你现在哪只牛需要最多的时间? 这一题可以这么看,回来的时候可以是从汇点(开会的地方)去到不同点的最短距离,然后从不同的点去的时候可以看做从汇点沿着路径的反方向看怎么走最短时间. 这样看这一题就简单很多了,因为没有负边,直接Dijkstra算法,算两次即可 #include <iostream> #include <f…

#include<iostream> #include<queue> #include<cstring> using namespace std; +,INF=0x3f3f3f3f; int h[N],e[N],ne[N],w[N],idx; int n,m,x; int dist[N]; bool st[N]; struct node{ int from; int to; int w; }p[N]; void add(int a,int b,int c) { e[id…

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: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…

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…

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: 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 …

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…