POJ2135:Farm Tour】的更多相关文章

题意:给定一个无向图,从1走到n再从n走回1,每个边只能走一遍,求最短路 题解:可以定义一个源点s,和一个汇点t s和1相连容量为2,费用为0, t和n相连容量为2,费用为0 然后所用的边的容量都定为1,跑一遍最小费用最大流即可 #include<cstdio> #include<cstdlib> #include<algorithm> #include<cstring> #include<queue> #include<vector>…
http://poj.org/problem?id=2135 题目大意: 从1到n再回来,每条边只能走一次,问最短路. —————————————————— 如果不告诉我是费用流打死不会想这个…… 我们把问题简化为1到n跑两遍,然后每条边容量为1,费用为长度. 然后建一个s和t,s到1容量为2,n到t容量为2. 跑费用流. (注意,本题是双向边-!) #include<cstdio> #include<iostream> #include<queue> #include…
题目链接:http://poj.org/problem?id=2135 Farm Tour Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 17672   Accepted: 6851 Description When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000…
  Farm Tour Time Limit: 2MS   Memory Limit: 65536KB   64bit IO Format: %I64d & %I64u Description When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of whic…
Farm Tour Time Limit: 1000ms Memory Limit: 65536KB This problem will be judged on PKU. Original ID: 2135 64-bit integer IO format: %lld      Java class name: Main   When FJ's friends visit him on the farm, he likes to show them around. His farm compr…
POJ 2135 Farm Tour (网络流,最小费用最大流) Description When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of which contains his house and the Nth of which contains the b…
Farm Tour 题目: 约翰有N块地,家在1号,而N号是个仓库.农场内有M条道路(双向的),道路i连接这ai号地和bi号地,长度为ci. 约翰希望依照从家里出发,经过若干地后达到仓库.然后再返回家中.假设要求往返不能经过同一条道路两次,求參观路线总长度最小值. 算法分析: 用最短路求解然后在删除第一次最短路中的边在求解一次最短路.这样是否可行?应该立即就能找到反例证明该方法不能总得到最优结果吧. 于是我们放弃把问题当作去和回的这样的想法,转而将问题当作求从1号顶点到N号顶点的两条没有公共边的…
Farm Tour Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 18150   Accepted: 7023 Description When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of…
Farm Tour Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 17230   Accepted: 6647 Description When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of…
题目:id=2135" target="_blank">poj 2135 Farm Tour 题意:给出一个无向图,问从 1 点到 n 点然后又回到一点总共的最短路. 分析:这个题目不读细致的话可能会当做最短路来做,最短路求出来的不一定是最优的,他是两条分别最短,但不一定是和最短. 我们能够用费用流来非常轻易的解决,建边容量为1,费用为边权.然后源点s连 1 .费用0 .容量 2 ,n点连接汇点,容量2,费用0,,就能够了. 注意这个题目是无向图,所以要建双向边. AC…
Farm Tour 题目描述 When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of which contains his house and the Nth of which contains the big barn. A total M (1 <= M…
Description When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of which contains his house and the Nth of which contains the big barn. A total M (1 <= M <…
解题关键:最小费用流 代码一:bellma-ford $O(FVE)$  bellman-ford求最短路,并在最短路上增广,速度较慢 #include<cstdio> #include<cstring> #include<algorithm> #include<cstdlib> #include<iostream> #include<cmath> #include<vector> #define inf 0x3f3f3f…
题目问的是从1到n再回到1边不重复走的最短路,本质是找1到n的两条路径不重复的尽量短的路. #include<cstdio> #include<cstring> #include<queue> #include<algorithm> using namespace std; #define INF (1<<30) #define MAXN 111 #define MAXM 2222 struct Edge{ int u,v,cap,cost,ne…
Description When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of which contains his house and the Nth of which contains the big barn. A total M (1 <= M <…
Description When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of which contains his house and the Nth of which contains the big barn. A total M (1 <= M <…
When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of which contains his house and the Nth of which contains the big barn. A total M (1 <= M <= 10000) pa…
题目大意: 给你一个n个农场,有m条道路,起点是1号农场,终点是n号农场,现在要求从1走到n,再从n走到1,要求不走重复路径,求最短路径长度. 算法讨论: 最小费用最大流.我们可以这样建模:既然要求不能走重复路,就相当于每条边的容量是1,我们只可以单向流过容量为1的流量.但是要注意,对于每一条边来说, 它可能是去路的边,也可能是回路的边,所以这个图是个无向图.在加边的时候,两个方向的边都要加.所以要加两组的边,流量为1像正常一样加边就可以了. 然后我们考虑,求这个“环”就是相当于求从1..N的两…
Description When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of which contains his house and the Nth of which contains the big barn. A total M (1 <= M <…
描述 When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of which contains his house and the Nth of which contains the big barn. A total M (1 <= M <= 10000)…
After the success of 2nd anniversary (take a look at problem FTOUR for more details), this 3rd year, Travel Agent SPOJ goes on with another discount tour. The tour will be held on ICPC island, a miraculous one on the Pacific Ocean. We list N places (…
题意: 有n个点和m条边,让你从1出发到n再从n回到1,不要求所有点都要经过,但是每条边只能走一次.边是无向边. 问最短的行走距离多少. 一开始看这题还没搞费用流,后来搞了搞再回来看,想了想建图不是很难,因为要保证每条边只能走一次,那么我们把边拆为两个点,一个起点和终点,容量是1,权重是这条路的长度.然后两个端点分别向起点连接容量是1权重是0的边,终点分别向两个端点连容量是1权重是0的边,从源点到1连容量为2权重为0的边,从n到汇点连容量为2权重为0的边. #include<stdio.h>…
题意:给一个无向图,FJ要从1号点出发到达n号点,再返回到1号点,但是路一旦走过了就会销毁(即回去不能经过),每条路长度不同,那么完成这趟旅行要走多长的路?(注:会有重边,点号无序,无向图!) 思路: 有重边,要用邻接表.所给的每条边都要变成4条有向边!否则可能一开始就到达不了终点了.最后要再加上一个源点和汇点,容量cap(源点,1)=2,指定只能走两次,再规定其他所给的边的容量是1就行了,当边被走过了,就自动增加了流,也就走不了了. 解释看代码更清晰. //#pragma comment(li…
题目大意:有一个无向图..农夫从1号点出发..要到达N号点..然后回到1号点..来回不能走相同的路径..问最短的距离是多少. 题解:又是不能走重复路径!基本图论算法直接扔掉上网络流.不能相同就边限1,然后就MCMF就行. 然后注意:MCMF无向边是要老老实实加两次的,不能向最大流一样偷偷换掉反向弧.…
Evacuation Plan Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 4914   Accepted: 1284   Special Judge Description The City has a number of municipal buildings and a number of fallout shelters that were build specially to hide municipal w…
http://www.spoj.com/problems/FTOUR2/en/ 题目大意:给一棵黑白染色的树,求边权和最大且经过黑点不超过K的路径. ———————————————————— 前排膜拜hzwer,借(抄)鉴(袭)了神犇的代码与思路,要看简洁的思路的话:http://hzwer.com/5984.html 然而……就算是抄了代码,还是没懂这题怎么做怎么办? 别着急,这道神神神题慢慢来. —————————————————————— 首先判断我们要放什么板子,很显然我们想到的是和树有…
两条路不能有重边,既每条边的容量是1.求流量为2的最小费用即可. //#pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<iostream> #include<sstream> #include…
题目链接 题意:无向图有N(N <= 1000)个节点,M(M <= 10000)条边:从节点1走到节点N再从N走回来,图中不能走同一条边,且图中可能出现重边,问最短距离之和为多少? 思路:很经典的构图(看题解的);每条原图中的边赋予cap为1,表示只走一次.超级源点s和汇点t分别和起点终点连边,cap为2,这里cap为2就直接限制了只能有两次最大流:同时最大流中以权值限制得到的就是最小费用:很注意的一点就是此题为无向图带权值,建图时每条有向边建成两条即总边数为4*M.由于spfa找最短路是有…
[题意]给出一张无向图,从1开始到n,求两条没有公共边的最短路,使得路程总和最小 每条边的权值设为费用,最大流量设为1,然后就是从源点到汇点流量为2的最小费用流. 因为是规定了流量,新建一个源点和一个汇点,源点到结点1连一条最大流量为2,费用为0的边,结点N到汇点连一条最大流量为2,费用为0的边,这样就规定好流量了. #include<cstdio> #include<cstring> #include<cmath> #include<iostream> #…
题目链接 给一个图, N个点, m条边, 每条边有权值, 从1走到n, 然后从n走到1, 一条路不能走两次,求最短路径. 如果(u, v)之间有边, 那么加边(u, v, 1, val), (v, u, 1, val), val是路的长度,代表费用, 1是流量. #include <iostream> #include <vector> #include <cstdio> #include <cstring> #include <algorithm&g…