题目链接:https://vjudge.net/problem/HDU-4185 Oil Skimming Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3016    Accepted Submission(s): 1262 Problem Description Thanks to a certain "green" re…

如下图:要求最多可以凑成多少对对象 大佬博客: https://blog.csdn.net/cillyb/article/details/55511666 https://blog.csdn.net/denghecsdn/article/details/77619308 https://www.cnblogs.com/wangjunyan/p/5563154.html 模板: int link[maxn],vis[maxn]; bool dfs(int x) { ; i <= num; i++)…

Oil Skimming Problem Description Thanks to a certain "green" resources company, there is a new profitable industry of oil skimming. There are large slicks of crude oil floating in the Gulf of Mexico just waiting to be scooped up by enterprising…

原文链接http://www.cnblogs.com/zhouzhendong/p/8231146.html 题目传送门 - HDU4185 题意概括 每次恰好覆盖相邻的两个#,不能重复,求最大覆盖次数.(引用大佬的http://blog.csdn.net/u011721440/article/details/38144339) 题解 我们对于每两个相邻#的建边(来回两条). 然后我们把格子黑白染色一下,发现黑点只会向白点连边,白点只会连向黑点连边,于是这个就是二分图了. 然后跑一跑匈牙利. 答…

<span style="font-family: Arial; font-size: 14.3999996185303px; line-height: 26px;">//题意,在一个N*N的矩阵里寻找最多有多少个</span><span style="font-size: 14px; line-height: 26px; font-family: 'Courier New', Courier, monospace; white-space: p…

Oil Skimming Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3903    Accepted Submission(s): 1616 题目链接:acm.hdu.edu.cn/showproblem.php?pid=4185 Description: Thanks to a certain "green" resou…

Problem Description Thanks to a certain "green" resources company, there is a new profitable industry of oil skimming. There are large slicks of crude oil floating in the Gulf of Mexico just waiting to be scooped up by enterprising oil barons. O…

Description Thanks to a certain "green" resources company, there is a new profitable industry of oil skimming. There are large slicks of crude oil floating in the Gulf of Mexico just waiting to be scooped up by enterprising oil barons. One such…

Oil Skimming Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3426    Accepted Submission(s): 1432 Problem Description Thanks to a certain "green" resources company, there is a new profitabl…

Oil Skimming Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 4185 Description Thanks to a certain "green" resources company, there is a new profitable industry of oil skimming. There ar…

题意大概是,海上漂浮着一些符号为#的石油,你要去搜集他们,但是你的勺子呢能且只能挖到两个单元的石油.问你最多能挖多少勺.注意 不能挖到纯净的海水,不然石油会被纯净的海水稀释的. 二分匹配,计算出里边有多少个'#',将这些编号,之后查找,如果他的周围是"#",就将这两个连一条边.之后跑一边匈牙利:点数就是'#'数目: Thanks to a certain "green" resources company, there is a new profitable ind…

题目大意:在一个N*N的矩阵里寻找最多有多少个“##”(横着竖着都行).     题目分析:重新构图,直接以相邻的两个油井算中间算以条边,然后进行匹配,看看两两之间最多能匹配多少对. #include<stdio.h> #include<string.h> #include<algorithm> #include<iostream> #include<vector> #include<queue> #include<cmath&…

题意:在大海里有一些石油 ‘#’表示石油, ‘.’表示水,有个人有一个工具可以回收这些石油,不过只能回收1*2大小的石油块,里面不能含有海水,要不就没办法使用了,求出来最多能回收多少块石油 分析:先把数据处理一下,给每一点石油都进行编号,然后查找一下四周联合是否能组成石油块,能的话就连接,因为一点有可能即在左边又在右边,所以最后的结果应该除去 2 ************************************************************************* #…

<题目链接> 题目大意: 给你一张图,图中有 '*' , '.' 两点,现在每次覆盖相邻的两个 '#' ,问最多能够覆盖几次. 解题分析: 无向图二分匹配的模板题,每个'#'点与周围四个方向的'#'建立匹配关系,然后用匈牙利跑一遍,因为匹配的两点各会进行相互匹配一次,所以最大匹配数为ans/2. #include <cstdio> #include <cstring> #include <algorithm> using namespace std; +;…

链接: http://acm.hdu.edu.cn/showproblem.php?pid=4185 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=82834#problem/G 与上题(H)相似 代码: #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std…

题目大意: 统计相邻(上下左右)的‘#’的对数. 解法: 与题目hdu1507 Uncle Tom's Inherited Land*类似,需要用奇偶建图.就是行+列为奇数的作为X集合,偶尔作为Y集合,都是‘#’就连边.最后求最大匹配. 数据有点大,直接建图会出错(我试过).可以按照‘#’出现的顺序给顶点编号. #include<iostream> #include<cstdio> #include<cstring> #include<queue> #inc…

传送门:hdu4185 Oil Skimming 题意:n*n的方格里有字符*和#,只能在字符#上放1*2的板子且不能相交,求最多能放多少个. 分析:直接给#字符编号,然后相邻的可以匹配,建边后无向图跑匈牙利算法,最后得到的最大匹配数/2. #include <cstdio> #include <cstring> #include <string> #include <cmath> #include <iostream> #include <…

链接: https://vjudge.net/problem/HDU-4185 题意: Thanks to a certain "green" resources company, there is a new profitable industry of oil skimming. There are large slicks of crude oil floating in the Gulf of Mexico just waiting to be scooped up by en…

Thanks to a certain "green" resources company, there is a new profitable industry of oil skimming. There are large slicks of crude oil floating in the Gulf of Mexico just waiting to be scooped up by enterprising oil barons. One such oil baron ha…

Oil Skimming Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2917    Accepted Submission(s): 1210 Problem Description Thanks to a certain "green" resources company, there is a new profitabl…

Oil Skimming Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Problem Description Thanks to a certain "green" resources company, there is a new profitable industry of oil skimming. There are large slicks of crude…

Current Time: 2016-03-11 17:45:36 Contest Type: Public Start Time: 2016-03-04 13:00:00 Contest Status: Running End Time: 2016-05-03 13:00:00 Manager: qwerqqq Clone this contest Edit Delete     ID Origin Title 1 / 1 Problem A HDU 1045 Fire Net 1 / 1 P…

    A-L 二分匹配 M-O 二分图多重匹配 P-Q 二分图最大权匹配 R-S 一般图匹配带花树 模板请自己找     ID Origin Title   61 / 72 Problem A HDU 1045 Fire Net   52 / 112 Problem B HDU 2444 The Accomodation of Students   45 / 86 Problem C HDU 1083 Courses   44 / 63 Problem D HDU 1281 棋盘游戏   35…

KUANGBIN带你飞 全专题整理 https://www.cnblogs.com/slzk/articles/7402292.html 专题一 简单搜索 POJ 1321 棋盘问题    //2019.3.18 POJ 2251 Dungeon Master POJ 3278 Catch That Cow  //4.8 POJ 3279 Fliptile POJ 1426 Find The Multiple  //4.8 POJ 3126 Prime Path POJ 3087 Shuffle…

题目链接:http://poj.org/problem?id=3020 题意:每个 ‘*’都需要一个1*2的东西覆盖,问最少需要多少个1*2的东西来覆盖这些‘*’ 和Oil Skimming的题解几乎一样 #include<stdio.h> #include<string.h> #include<iostream> #include<algorithm> #include<queue> using namespace std; #define N…

二分匹配:二分图的一些性质 二分图又称作二部图,是图论中的一种特殊模型. 设G=(V,E)是一个无向图,如果顶点V可分割为两个互不相交的子集(A,B),并且图中的每条边(i,j)所关联的两个顶点i和j分别属于这两个不同的顶点集(i in A,j in B),则称图G为一个二分图. 1.一个二分图中的最大匹配数等于这个图中的最小点覆盖数 König定理是一个二分图中很重要的定理,它的意思是,一个二分图中的最大匹配数等于这个图中的最小点覆盖数.如果你还不知道什么是最小点覆盖,我也在这里说一下:假如选…

Oil Skimming Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1487    Accepted Submission(s): 612 Problem Description Thanks to a certain "green" resources company, there is a new profitable…

[kuangbin带你飞]专题1-23 专题一 简单搜索 POJ 1321 棋盘问题POJ 2251 Dungeon MasterPOJ 3278 Catch That CowPOJ 3279 FliptilePOJ 1426 Find The MultiplePOJ 3126 Prime PathPOJ 3087 Shuffle'm UpPOJ 3414 PotsFZU 2150 Fire GameUVA 11624 Fire!POJ 3984 迷宫问题HDU 1241 Oil Deposit…

专题一 简单搜索 POJ 1321 棋盘问题POJ 2251 Dungeon MasterPOJ 3278 Catch That CowPOJ 3279 FliptilePOJ 1426 Find The MultiplePOJ 3126 Prime PathPOJ 3087 Shuffle'm UpPOJ 3414 PotsFZU 2150 Fire GameUVA 11624 Fire!POJ 3984 迷宫问题HDU 1241 Oil DepositsHDU 1495 非常可乐HDU 26…

题目链接:Oil 感觉同时几线作战有点吃不消啊-- 这道题有一个显然的结论,那就是最优的直线一定过某条线段的端点. 仔细想想很有道理.如果最终的直线没有过线段的端点的话,那么这条直线就一定可以平移,直到过端点为止. 于是我们可以枚举直线上的一个点,由于直线不能与线段平行,那么与枚举的点纵坐标不同的线段就对应着一个斜率区间.于是这个问题就转化成了一个经典问题:有$n$个区间,第$i$个区间$[l_i,r_i]$会给区间内的所有位置(可以不是整数)加上一个权值$c_i$,求最后所有位置中最大的权值.…