Farm Tour POJ - 2135 (最小费用流)】的更多相关文章

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…
题目链接:http://poj.org/problem?id=2135 今天学习最小费用流.模板手敲了一遍. 产生了一个新的问题:对于一条无向边,这样修改了正向边容量后,反向边不用管吗? 后来想了想,得出了个结论.路径所选的边只会包括正反中的一条. #include <iostream> #include <cstdio> #include <cstring> #include <vector> #include <queue> #include…
题目大意: 给定一个n个点m条边的无向图 求从点1去点n再从点n回点1的不重叠(同一条边不能走两次)的最短路 挑战P239 求去和回的两条最短路很难保证不重叠 直接当做是由1去n的两条不重叠的最短路 这样就变成了由1去n流量为2的最小费用流 #include <bits/stdc++.h> #define INF 0x3f3f3f3f using namespace std; ; int n,m; struct EDGE { int v,w,c,r; }; vector <EDGE>…
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…
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 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…
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…
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: 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…
题目链接:http://poj.org/problem?id=2135 Farm Tour Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 14821   Accepted: 5657 Description When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000…
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: 65536K Total Submissions: 10840   Accepted: 4011 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…
题目链接 Farm Tour Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 18961   Accepted: 7326 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 fir…
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…
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: 19207   Accepted: 7441 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…
B - Sightseeing tour POJ - 1637 https://blog.csdn.net/qq_36551189/article/details/80905345 首先要了解一下欧拉回路的基本思路. 欧拉回路:如果是无向图,那么每一个点连的边的数量为偶数,如果是有向图,那么每一个点的入度要等于出度. 欧拉路径:这个欧拉路径是没有成环的,如果是无向图,那么除了两个点连的边是奇数,其他都是偶数, 如果是有向图,那么除了有一个点入度比出度大1,有一个点的出度比入度大1 ,其他都是入度…
[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> #…
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…
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…
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 <…
[题目链接] http://poj.org/problem?id=2135 [题目大意] 有一张无向图,求从1到n然后又回来的最短路 同一条路只能走一次 [题解] 题目等价于求从1到n的两条路,使得两条路的总长最短 那么就等价于求总流量为2的费用流 [代码] #include <cstdio> #include <cstring> #include <algorithm> #include <vector> #include <queue> #i…
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.…
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…
累了就要写题解,近期总是被虐到没脾气. 来回最短路问题貌似也能够用DP来搞.只是拿费用流还是非常方便的. 能够转化成求满流为2 的最小花费.一般做法为拆点,对于 i 拆为2*i 和 2*i+1.然后连一条流量为1(花费依据题意来定) 的边来控制每一个点仅仅能通过一次. 额外加入source和sink来控制满流为2. 代码都雷同,以HDU3376为例. #include <algorithm> #include <iostream> #include <cstring>…
inf开太小错了好久--下次还是要用0x7fffffff #include<stdio.h> #include<string.h> #include<vector> #include<queue> #include<algorithm> using namespace std; const int N=5024; const int inf=0x7fffffff; struct Edge { int from,to,cap,flow,cost;…