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 vertically, to an adjacent point. For each little man, you need to pay a $1 travel fee for every step he moves, unt…

就是一个简单题 四个月前a的一道题,今天又看到了,再a一遍吧. 好吧 我想多了 用了bfs求最短路  其实不用的 因为没有障碍物 #include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <cctype> #include <set> #include <vector>…

题意:有N个供应商,M个店主,K种物品.每个供应商对每种物品的的供应量已知,每个店主对每种物品的需求量的已知,从不同的供应商运送不同的货物到不同的店主手上需要不同的花费,又已知从供应商m送第k种货物的单位数量到店主n手上所需的单位花费.供应是否满足需求?如果满足,最小运费是多少? 思路:这题一读完就知道是费用流了,刚开始想着拆点,不过算了一下,把m个供应商拆成m*k个点,n个店主拆成n*k个点,加起来有5000多个点,肯定会超时的,看了网上说每种商品求一次费用流就可以了,就是100个点求50次.…

题意抽象出来就是给了一个费用流的残存网络,判断该方案是不是最优方案,如果不是,还要求给出一个更优方案. 在给定残存网络上检查是否存在负环即可判断是否最优. 沿负环增广一轮即可得到更优方案. 考虑到制作模板的需要,我用前向星建图做的.此题用邻接矩阵应该更方便. #include<queue> #include<cstdio> #include<algorithm> #include<cstring> #define rep(i,a,b) for(int i=a…

题意: 1.一个人从[1,1] ->[n,n] ->[1,1] 2.仅仅能走最短路 3.走过的点不能再走 问最大和. 对每一个点拆点限流为1就可以满足3. 费用流流量为2满足1 最大费用流,先给图取负,结果再取负,满足2 #include <stdio.h> #include <string.h> #include <iostream> #include <math.h> #include <queue> #include <s…

Lunch Time http://acm.hdu.edu.cn/showproblem.php?pid=4807 Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 782    Accepted Submission(s): 183 Problem Description The campus of Nanjing University…

Coding Contest http://acm.hdu.edu.cn/showproblem.php?pid=5988 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 5337    Accepted Submission(s): 1256 Problem Description A coding contest will be he…

A Plug for UNIX Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 15597   Accepted: 5308 Description You are in charge of setting up the press room for the inaugural meeting of the United Nations Internet eXecutive (UNIX), which has an int…

The Windy's Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 5362   Accepted: 2249 Description The Windy's is a world famous toy factory that owns M top-class workshop to make toys. This year the manager receives N orders for toys. The ma…

2018南京I题: dinic,链式前向星,数组队列,当前弧优化,不memset全部数组,抛弃满流点,bfs只找一条增广路,每次多路增广 #include <bits/stdc++.h> #define ll long long #define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0) #define rep(ii,a,b) for(int ii=a;ii<=b;++ii) using namespace std; con…