Pahom on Water Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 885    Accepted Submission(s): 409 Problem Description Pahom on Water is an interactive computer game inspired by a short story of…

[解题报告] Leapin' Lizards HDU 2732 网络流 题外话 在正式讲这个题目之前我想先说几件事 1. 如果大家要做网络流的题目,我在网上看到一个家伙,他那里列出了一堆网络流的题目,而且还给他们分门别类,并且标注了难度,感觉挺好的,网址是[夏天的风](https://blog.csdn.net/shahdza/article/details/7779537) 2. 这道题有个坑点!!!! 正题 首先,当然是直接贴上[题目](http://acm.hdu.edu.cn/showp…

http://acm.hdu.edu.cn/showproblem.php?pid=1083 二分图匹配用得很多 这道题只需要简化的二分匹配 #include<iostream> #include<cstdio> #include<cstring> #define maxm 410 using namespace std; int p,n; int master[maxm]; int linking[maxm][maxm]; int has[maxm]; int sol…

Poroblem Redraw Beautiful Drawings (HDU4888) 题目大意 一个n行m列的矩形,只能填0~k的数字. 给定各行各列的数字和,判定有无合法的方案数.一解给出方案,多解输出给定字符串. 解题分析 一个经典的网络流建图. 由S向行连流量为该行数字和的边,由列向T连流量为该列数字和的边,从行向列连流量为k的边. 若满流说明有解. 在残余网络中从每个点开始dfs,若找到一个点数大于2的环,说明有多解. 参考程序 #include <cstdio> #include…

裸的网络流,递归的dinic会爆栈,在第一行加一句就行了 #pragma comment(linker, "/STACK:1024000000,1024000000") #include <iostream> #include <stdio.h> #include <string.h> #include <cstring> #include <algorithm> #include <vector> #define…

职务地址:HDU 4292 水题. 因为每一个人仅仅能有1份,所以须要拆点限制流量.建图方法为,建一源点与汇点.将食物与源点相连,权值为食物额数量,将饮料与汇点相连,权值为饮料数量..然后将人进行拆点为i和i',将相应的i与i'连边权值为1.将i与它所相应的YES的食物连边.将i'与它所相应的YES的饮料连边.一次求最大流. 代码例如以下: #include <iostream> #include <stdio.h> #include <string.h> #inclu…

http://acm.hdu.edu.cn/showproblem.php?pid=4292 给一些人想要的食物和饮料,和你拥有的数量,问最多多少人可以同时获得一份食物和一份饮料 写的时候一共用了2种拆点建图的方法... 1. 起点连接饮料和食物,容量为拥有的数量 每个人被拆成三个点,$a,b,c$  $a$被想要的食物连接,$b$被想要的饮料连接,$c$被$a,b$连接,容量均为1 然后$c$点连接汇点,容量为2,最后遍历所有静态链表节点,对于所有指向汇点的边,如果剩余容量为0,则答案++ 然…

方格取数(1) Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7717    Accepted Submission(s): 2911 Problem Description 给你一个n*n的格子的棋盘,每个格子里面有一个非负数.从中取出若干个数,使得任意的两个数所在的格子没有公共边,就是说所取的数所在的2个格子不能相邻,并且取出的数…

题目大意 有n个灯,m个开关,由于线路乱接导致可能有多个开关对应一个灯(并联),有的灯在开关开的时候亮 有的灯在开关关的时候亮,[每个开关最多对应两盏灯],试找出一种开关的ON,OFF状态,使得所有灯都亮. (注意不要漏读黑框内的内容) 解法1:网络流 对于一个开关有一下几种情况: 1.开关只连向一个灯,直接设定开关点亮此灯. 2.开关连向两个开关状态需求相同的灯,直接把开关拨到相应状态,点亮两个灯. 3.开关连向两个需求不同冲突的灯,(只能选择两者之一点亮). 接下来网络流即可[注意区分n,m…

如果我们先手拿完所有苹果再去考虑花费的话. S -> 摄像头 -> 苹果 -> T 就相当于找到一个最小割使得S和T分开. ans = sum - flow. 然后对于这一个模型, 我们可以不用网络流去解决. 我们从叶子出发,然后从下往上合并. 每次到一个节点的时候,我们先把摄像机所对应的影响去除. 然后把这个点的剩下流量传给父亲. 代码: #include<bits/stdc++.h> using namespace std; #define Fopen freopen(&…

题目链接:https://vjudge.net/contest/299467#problem/K 这个题目从数据范围来看可以发现是网络流,怎么建图呢?这个其实不是特别难,主要是读题难. 这个建图就是把源点和每一个蜥蜴存在的点相连,汇点和可以跑出去的相连,因为这个题目对于每一个点都有次数要求,所以就要拆点. #include <cstdio> #include <cstdlib> #include <cstring> #include <string> #in…

https://vjudge.net/problem/HDU-4183 题意: 这道题目的英文实在是很难理解啊. 给出n个圆,每个圆有频率,x.y轴和半径r4个属性,每次将频率为400的圆作为起点,频率为789点作为终点.从源点到汇点时必须从频率小的到频率大的,而从汇点到源点时必须从频率大的到频率小的.前提时这两个圆必须严格相交.每个点只能走一次.判断是否能从起点出发到达终点,并再次返回起点. 思路: 其实就是判断最大流是否大于等于2.因为每个点只能走一次,用拆点法. #include<iost…

给出一个有向图,以及src和dst.判断是否存在从src到dst的两条路径,使得除了src和dst外,没有其它点同时属于两条路径. 给每个点一个为1的点容量(src和dst为2),边的容量也是1,然后判断最大流是否大于等于2. 收获: 边不能重复:将点拆成两个点考虑,然后考虑匹配. 点不能重复:给每个点一个点容量(其实还是拆点),然后考虑流. #include <cstdio> #include <cstring> #include <queue> #include &…

Control Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2247    Accepted Submission(s): 940 Problem Description You, the head of Department of Security, recently received a top-secret informatio…

欢迎参加——每周六晚的BestCoder(有米!) Pahom on Water Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 678    Accepted Submission(s): 312 Problem Description Pahom on Water is an interactive computer game ins…

Steady Cow Assignment Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6422   Accepted: 2202 Description Farmer John's N (1 <= N <= 1000) cows each reside in one of B (1 <= B <= 20) barns which, of course, have limited capacity. So…

Pahom on Water Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 629    Accepted Submission(s): 288 Problem Description Pahom on Water is an interactive computer game inspired by a short story of…

Ford-Fulkerson方法依赖于三种重要思想,这三个思想就是:残留网络,增广路径和割. Ford-Fulkerson方法是一种迭代的方法.开始时,对所有的u,v∈V有f(u,v)=0,即初始状态时流的值为0.在每次迭代中,可通过寻找一条“增广路 径”来增加流值.增广路径可以看成是从源点s到汇点t之间的一条路径,沿该路径可以压入更多的流,从而增加流的值.反复进行这一过程,直至增广路 径都被找出来,根据最大流最小割定理,当不包含增广路径时,f是G中的一个最大流. #include <stdio…

题目   You, a part-time dining service worker in your college’s dining hall, are now confused with a new problem: serve as many people as possible.  The issue comes up as people in your college are more and more difficult to serve with meal: They eat o…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2883 Almost everyone likes kebabs nowadays (Here a kebab means pieces of meat grilled on a long thin stick). Have you, however, considered about the hardship of a kebab roaster while enjoying the delicio…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3572 Our geometry princess XMM has stoped her study in computational geometry to concentrate on her newly opened factory. Her factory has introduced M new machines in order to process the coming N tasks.…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3549 Network flow is a well-known difficult problem for ACMers. Given a graph, your task is to find out the maximum flow for the weighted directed graph. Input The first line of input contains an integer…

题目地址:HDU 3468 这道题的关键在于能想到用网络流.然后还要想到用bfs来标记最短路中的点. 首先标记方法是,对每个集合点跑一次bfs,记录全部点到该点的最短距离.然后对于随意一对起始点来说,仅仅要这个点到起点的最短距离+该点到终点的最短距离==起点到终点的最短距离,就说明这点在某条从起点到终点的最短路上. 然后以集合点建X集,宝物点建Y集构造二分图,将从某集合点出发的最短路中经过宝物点与该集合点连边.剩下的用二分匹配算法或最大流算法都能够.(为什么我的最大流比二分匹配跑的还要快....…

传送门:pid=4888">[HDU]4888 Redraw Beautiful Drawings 题目分析: 比赛的时候看出是个网络流,可是没有敲出来.各种反面样例推倒自己(究其原因是不愿意写暴力推断的).. 首先是简单的行列建边.源点向行建边.容量为该行元素和,汇点和列建边.容量为该列元素和.全部的行向全部的列建边,容量为K. 跑一次最大流.满流则有解,否则无解. 接下来是推断解是否唯一. 这个题解压根没看懂.还是暴力大法好. 最简单的思想就是枚举在一个矩形的四个端点.设A.D为主对角…

HDU 3416 Marriage Match IV (最短路径,网络流,最大流) Description Do not sincere non-interference. Like that show, now starvae also take part in a show, but it take place between city A and B. Starvae is in city A and girls are in city B. Every time starvae can…

HDU 3081 Marriage Match II (网络流,最大流,二分,并查集) Description Presumably, you all have known the question of stable marriage match. A girl will choose a boy; it is similar as the game of playing house we used to play when we are kids. What a happy time as…

HDU 3605 Escape (网络流,最大流,位运算压缩) Description 2012 If this is the end of the world how to do? I do not know how. But now scientists have found that some stars, who can live, but some people do not fit to live some of the planet. Now scientists want you…

HDU 3338 Kakuro Extension (网络流,最大流) Description If you solved problem like this, forget it.Because you need to use a completely different algorithm to solve the following one. Kakuro puzzle is played on a grid of "black" and "white" ce…

POJ 2711 Leapin' Lizards / HDU 2732 Leapin' Lizards / BZOJ 1066 [SCOI2007]蜥蜴(网络流,最大流) Description Your platoon of wandering lizards has entered a strange room in the labyrinth you are exploring. As you are looking around for hidden treasures, one of…

HDU 4289 Control (网络流,最大流) Description You, the head of Department of Security, recently received a top-secret information that a group of terrorists is planning to transport some WMD(Weapon of Mass Destruction)from one city (the source) to another o…