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…

Going Home Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 15944   Accepted: 8167 Description On a grid map there are n little men and n houses. In each unit time, every little man can move one unit step, either horizontally, or vertical…

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…

图论算法-最小费用最大流模板[EK;Dinic] EK模板 const int inf=1000000000; int n,m,s,t; struct node{int v,w,c;}; vector<node> map[10010]; int flow[10010][10010]; bool inq[10010]; int d[10010]; int pre[10010],pref[10010]; int minc,maxf; int main() { cin>>n>>…

Matrix Again Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)Total Submission(s): 4255    Accepted Submission(s): 1233 Problem Description Starvae very like play a number game in the n*n Matrix. A positive integer…

https://www.luogu.org/problem/P3381 题目描述 如题,给出一个网络图,以及其源点和汇点,每条边已知其最大流量和单位流量费用,求出其网络最大流和在最大流情况下的最小费用. 输入格式 第一行包含四个正整数N.M.S.T,分别表示点的个数.有向边的个数.源点序号.汇点序号. 接下来M行每行包含四个正整数ui.vi.wi.fi,表示第i条有向边从ui出发,到达vi,边权为wi(即该边最大流量为wi),单位流量的费用为fi. 输出格式 一行,包含两个整数,依次为最大流量和…

模板从  这里   搬运,链接博客还有很多网络流题集题解参考. 最大流模板 ( 可处理重边 ) ; const int INF = 0x3f3f3f3f; struct Edge { int from,to,cap,flow; Edge(){} Edge(int from,int to,int cap,int flow):from(from),to(to),cap(cap),flow(flow){} }; struct Dinic { int n,m,s,t; //结点数,边数(包括反向弧),源…

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…

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…