POJ 3268 (dijkstra变形)】的更多相关文章

题目链接 :http://poj.org/problem?id=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 be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roa…
思路:正向建边,一遍Dijkstra,反向建边,再一遍Dijkstra.ans加在一起输出最大值. (SPFA也行--) // by SiriusRen #include <queue> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; #define N 1005 int n,m,X,tot=0,maxx=0,first[N],v[N*N],w[N*…
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…
POJ.1797 Heavy Transportation (Dijkstra变形) 题意分析 给出n个点,m条边的城市网络,其中 x y d 代表由x到y(或由y到x)的公路所能承受的最大重量为d,求从1到n的所有通路中,所能经过的的最大重量的车为多少. 2. 代码总览 #include <cstdio> #include <cstring> #include <algorithm> #include <queue> #include <stack&…
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…
题目链接: http://acm.nyist.net/JudgeOnline/problem.php?pid=1248 描述 神秘的海洋,惊险的探险之路,打捞海底宝藏,激烈的海战,海盗劫富等等.加勒比海盗,你知道吧?杰克船长驾驶着自己的的战船黑珍珠1号要征服各个海岛的海盜,最后成为海盗王. 这是一个由海洋.岛屿和海盗组成的危险世界.杰克船长准备从自己所占领的岛屿A开始征程,逐个去占领每一个岛屿.面对危险重重的海洋与诡谲的对手,如何凭借智慧与运气,建立起一个强大的海盗帝国. 杰克船长手头有一张整个…
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…
light1002:传送门 [题目大意] n个点m条边,给一个源点,找出源点到其他点的‘最短路’ 定义:找出每条通路中最大的cost,这些最大的cost中找出一个最小的即为‘最短路’,dijkstra变形.dis[i]为s->i的‘最短路’ #include<bits/stdc++.h> ][],dis[],vis[]; using namespace std; int n; void dij(int s) { int i,j,k; ; i<n; i++) { dis[i]=mp[…
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…