建图           源点  ->     每个人  ->           每段时间      ->      汇点 时间要离散化一下 分成一些时间段 权                     inf    ti[i]*(time[i]-time[i-1])  m*(time[i]-time[i-1]) n人 m个机器 开始时间  总共要的数目 底线时间  每个所需时间 如果可以达到输出Y 否则N #include<stdio.h> #include<algo…

HDU 2883 kebab 题目链接 题意:有一个烧烤机,每次最多能烤 m 块肉.如今有 n 个人来买烤肉,每一个人到达时间为 si.离开时间为 ei,点的烤肉数量为 ci,每一个烤肉所需烘烤时间为 di.注意一个烤肉能够切成几份来烤 思路:把区间每一个点存起来排序后.得到最多2 * n - 1个区间,这些就表示几个互相不干扰的时间,每一个时间内仅仅可能有一个任务器做.这样建模就简单了.源点连向汇点,容量为任务须要总时间,区间连向汇点,容量为区间长度.然后每一个任务假设包括了某个区间,之间就连…

网络流 HDU 3549 Flow Problem 题目:pid=3549">http://acm.hdu.edu.cn/showproblem.php?pid=3549 用增广路算法进行求解.注意的问题有两个: 1. 每次增广的时候,反向流量也要进行更行.一開始没注意,WA了几次 ORZ 2. 对于输入的数据,容量要进行累加更新. // 邻接矩阵存储 #include <bits/stdc++.h> using namespace std; const int INF = 0…

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…

Problem Description 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 delicious food? Well, here's a chance f…

kebab Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1243    Accepted Submission(s): 516 Problem Description Almost everyone likes kebabs nowadays (Here a kebab means pieces of meat grilled on…

kebab Time Limit: 1000ms Memory Limit: 32768KB This problem will be judged on HDU. Original ID: 288364-bit integer IO format: %I64d      Java class name: Main Almost everyone likes kebabs nowadays (Here a kebab means pieces of meat grilled on a long…

建图   源点    ->   1024类人   ->   星球   ->     汇点 权             每类人数目       星球容量     星球容量 列举 0~1024  一位是1 那么和对应的星球建边 #include<stdio.h> #include<algorithm> #include<string.h> #include<queue> #include<math.h> using namespac…

题意: 有n个人去撸串,每个人都能决定自己的串上有几块肉,每一块肉都要花费一个单位时间才熟,烤炉一次能烤m块肉 给出每个人的起始时间.终止时间.要几串.每个串上有几块肉,问能否满足所有的人 (啥?题不是这么说的?...这样看就对了...) 解析: 和hdu3572一样 只不过这个时间的范围比较大 所以就不能以时间点建点了  所以要以时间段建点 把所有的时间放到数组里,这里我用的是vector,然后排序,去重,对于每个人 遍历所有的时间,如果有个时间段在当前人购买的时间里  就把人和这个时间段连边…

kebab Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1603    Accepted Submission(s): 677 Problem Description Almost everyone likes kebabs nowadays (Here a kebab means pieces of meat grilled on…

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 delicious food? Well, here's a chance for you to help the p…

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…

HDU 4292 Food (网络流,最大流) Description 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…

HDU 4280 Island Transport(网络流,最大流) Description In the vast waters far far away, there are many islands. People are living on the islands, and all the transport among the islands relies on the ships. You have a transportation company there. Some route…

POJ 1087 A Plug for UNIX / HDU 1526 A Plug for UNIX / ZOJ 1157 A Plug for UNIX / UVA 753 A Plug for UNIX / UVAlive 5418 A Plug for UNIX / SCU 1671 A Plug for UNIX (网络流) Description You are in charge of setting up the press room for the inaugural meet…

题目链接:https://cn.vjudge.net/problem/HDU-1565 Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) 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,并且取出的数的和最大.   Input 包括多个测试实例,每…

[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=3338 题意:在一个n*m的地图里面,有黑方块和白方块,黑方块可能是“XXXXXXX”或者“YYY/YYY”,这里的YYY代表可能为数字,如果是在“/”左边出现数字,代表在它下面的该列的白方块的和加起来要等于这个数字,如果是在“/”右边出现数字,代表它右边的该行的白方块的和加起来要等于这个数字.我们要做的就是求出这些白方块上的数字,并按照要求输出. 思路:看完题意一脸懵逼,想了一个下午还是不知道怎么写.无奈只能…

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…

题目大意:有一个类似于迷宫搜索的图,‘.’代表的是无人的路,'X'代表有人的点,'#'代表此点不可通过,'@'代表门口.每个位置每一秒钟只能站一个人,每个位置到上下左右点的时间为1,问你所有人能不能出去,能出去输出所有人都出去的最小时间,否则输出-1. 链接:点我 A: 增加源点src,和汇点dest,然后根据每个时间点建出分层图,每个时间对应一层,对于每层图的构造如下 B:给每个格子标 上号Xi, 由于每个格子一次只能占一人,所以把每个格子分为两个点xa,xb,连上容量为1的有向边,对于格子为…

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…