Fire Net Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7998    Accepted Submission(s): 4573 Problem Description Suppose that we have a square city with straight streets. A map of a city is a s…

题意:任意两个个'车'不能出现在同一行或同一列,当然如果他们中间有墙的话那就没有什么事,问最多能放多少个'车' 代码+注释: 1 //二分图最大匹配问题 2 //难点在建图方面,如果这个图里面一道墙也没有,那么可以说就是在横坐标(1...n)和纵坐标(1...n) 3 //中个挑选出来一个,那个点就是我们要放置炮台的地方,么个点只能用一次,这样就能保证正确性 4 //现在现在里面有墙了,在横坐标方面那么看到一个墙就把横坐标的选择数加1,每次换行也加1 5 //纵坐标也做相应处理,那么我们就可以得…

题意:n*n的棋盘上放置房子.同一方同一列不能有两个,除非他们之间被墙隔开,这种话. 把原始图分别按行和列缩点 建图:横竖分区.先看每一列.同一列相连的空地同一时候看成一个点,显然这种区域不可以同一时候放两个点. 这些点作为二分图的X部.同理在对全部的 行用同样的方法缩点.作为Y部. #include<cstdio> #include<iostream> #include<algorithm> #include<cmath> #include<set&…

HDOJ(HDU).1045 Fire Net [从零开始DFS(7)] 点我挑战题目 从零开始DFS HDOJ.1342 Lotto [从零开始DFS(0)] - DFS思想与框架/双重DFS HDOJ.1010 Tempter of the Bone [从零开始DFS(1)] -DFS四向搜索/奇偶剪枝 HDOJ(HDU).1015 Safecracker [从零开始DFS(2)] -DFS四向搜索变种 HDOJ(HDU).1016 Prime Ring Problem (DFS) [从零开…

Description Suppose that we have a square city with straight streets. A map of a city is a square board with n rows and n columns, each representing a street or a piece of wall. A blockhouse is a small castle that has four openings through which to s…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=1045 Fire Net Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 15715    Accepted Submission(s): 9519 Problem Description Suppose that we have a square…

<题目链接> 题目大意: 这题意思是给出一张图,图中'X'表示wall,'.'表示空地,可以放置炮台,同一条直线上只能有一个炮台,除非有'X'隔开,问在给出的图中最多能放置多少个炮台. 解题分析: 本题可用DFS求解 >>> ,但是二分匹配的想法更加巧妙,效率也更高.二分匹配的主要思想就是,对矩阵的行连通块和列连通块进行标号,然后根据矩阵的每个点,建立对应的行连通块和列连通块之间的待匹配关系,然后利用匈牙利进行正式匹配,这样当某个行联通块与某个列连通块正式确立匹配关系的时候,…

http://acm.hdu.edu.cn/showproblem.php?pid=1045 Fire Net Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1045 Description Suppose that we have a square city with straight streets. A map of a city…

以防万一,题目原文和链接均附在文末.那么先是题目分析: [一句话题意] 给定大小的棋盘中部分格子存在可以阻止互相攻击的墙,问棋盘中可以放置最多多少个可以横纵攻击炮塔. [题目分析] 这题本来在搜索专题里出现的..这回又在二分查找匹配专题出现了..所以当然要按照二分匹配的方法解而不是爆搜(虽然爆搜能过). 问题主要就是如何缩点建图.为了使得blockhouse不能互相攻击,那么使用每行的相邻的点缩点,每列的相邻的点缩点,连边的条件就是两个点存在有相交的部分,最后这两组点求最大匹配就行了. [算法流…

Fire Net Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5863    Accepted Submission(s): 3280 Problem Description Suppose that we have a square city with straight streets. A map of a city is a s…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1281 棋盘游戏 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2905    Accepted Submission(s): 1702 Problem Description 小希和Gardon在玩一个游戏:对一个N*M的棋盘,在格子里放尽…

传送门:http://acm.hdu.edu.cn/showproblem.php?pid=1045 Fire Net Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 14670    Accepted Submission(s): 8861 Problem Description Suppose that we have a squar…

Cyclic Tour Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/65535 K (Java/Others)Total Submission(s): 1197    Accepted Submission(s): 626 Problem Description There are N cities in our country, and M one-way roads connecting them. Now Li…

Fire Net Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1045 Description Suppose that we have a square city with straight streets. A map of a city is a square board with n rows and n columns, e…

Fire Net Time Limit: 2 Seconds      Memory Limit: 65536 KB Suppose that we have a square city with straight streets. A map of a city is a square board with n rows and n columns, each representing a street or a piece of wall. A blockhouse is a small c…

原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=1045 Fire Net Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8073    Accepted Submission(s): 4626 Problem Description Suppose that we have a squar…

HDU 1045 题意: 在一个n*n地图中,有许多可以挡住子弹的墙,问最多可以放几个炮台,使得炮台不会相互损害.炮台会向四面发射子弹. 思路: 把行列分开做,先处理行,把同一行中相互联通的点缩成一个点.再处理列,同样缩成一个点.然后把行列中,交点不是墙的点连一条边.对这个图跑网络流或者二分图匹配即可. #include <algorithm> #include <iterator> #include <iostream> #include <cstring>…

//是象棋里的车 符合二分匹配 # include<stdio.h> # include<algorithm> # include<string.h> using namespace std; int n,m,pp[110][110],map[110],vis[110]; int bfs(int x) { for(int i=1;i<=m;i++) { if(!vis[i]&&pp[x][i]) { vis[i]=1; if(!map[i]||bf…

题目大意: 这个是以前做过的一道DFS题目,当时是完全暴力写的. 给你一个N代表是N*N的矩阵,矩阵内 ‘X’代表墙, ‘.’代表通道. 问这个矩阵内最多可以放几个碉堡, 碉堡不能在同一行或者同一列,除非他们中间有墙.   二分图做法思想:我们用行去匹配列,判断最大匹配数. 我们需要重新构图, 假如一行中 (  ..X..X.. ) 那么在这一行中我们其实是可以分割到三个不同的行(因为中间隔有X).然后对这个三个行进行编号.同理列也是一样的.当我们完全构好图后就可以做完全匹配了,其他的跟HDU…

题目链接:http://acm.split.hdu.edu.cn/showproblem.php?pid=1045 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Problem Description Suppose that we have a square city with straight streets. A map of a city is a square board…

http://acm.split.hdu.edu.cn/showproblem.php?pid=1045 http://acm.hdu.edu.cn/showproblem.php?pid=1045 题目链接(总有一个可以点开)…… 题意:给出一张地图,上面只有两种字符(.和X),X相当于无法穿透的墙.问这张地图上最多可以放置多少个互不攻击的子弹…… 第一反应是搜索,然而后来看到网上有些大神说这是二分图匹配……看来还是要研究下. 我剪枝的思路:对每个点设置所有关联的点(上下左右四个方向直到被墙挡…

题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=1045 题目大意为给定一个最大为4*4的棋盘,棋盘可以放置堡垒,处在同一行或者同一列的堡垒可以相互攻击,在棋盘上也有城墙,可以隔离同一行同一列堡垒的相互攻击,"X"表示有城墙," . "表示空地,问该棋盘最多可以放置多少个堡垒? 此题为二分图匹配问题,但是起初我不知道怎么建图,看题解发现一个规律,同一行同一列的连续空地不能放置两个堡垒,那么同一行同一列的连续空地的交点放置…

Description Suppose that we have a square city with straight streets. A map of a city is a square board with n rows and n columns, each representing a street or a piece of wall. A blockhouse is a small castle that has four openings through which to s…

Fire Net Problem Description Suppose that we have a square city with straight streets. A map of a city is a square board with n rows and n columns, each representing a street or a piece of wall. A blockhouse is a small castle that has four openings t…

Consider a town where all the streets are one-way and each street leads from one intersection to another. It is also known that starting from an intersection and walking through town's streets you can never reach the same intersection i.e. the town's…

The Accomodation of Students Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8418    Accepted Submission(s): 3709 Problem Description There are a group of students. Some of them may know each ot…

Fire Net Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 14780    Accepted Submission(s): 8932 Problem Description Suppose that we have a square city with straight streets. A map of a city is a…

(点击此处查看原题) 匈牙利算法简介 个人认为这个算法是一种贪心+暴力的算法,对于二分图的两部X和Y,记x为X部一点,y为Y部一点,我们枚举X的每个点x,如果Y部存在匹配的点y并且y没有被其他的x匹配,那就直接匹配:如果Y中已经没有可以和x匹配的点(包括可以匹配的点已经被其他的x匹配),那就让已经匹配的y的原配x'寻找其他可以匹配的y’,并将y和x匹配,最后,统计出匹配的对数 (详细了解的话,可以看看这位的博客:https://blog.csdn.net/sunny_hun/article/de…

Fire Net Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8239    Accepted Submission(s): 4734 Problem Description Suppose that we have a square city with straight streets. A map of a city is a s…

题意:交换任意两行或两列,使主对角线全为1. 分析: 1.主对角线都为1,可知最终,第一行与第一列匹配,第二行与第二列匹配,……. 2.根据初始给定的矩阵,若Aij = 1,则说明第i行与第j列匹配,据此求最大匹配数cnt,若cnt==N,才可通过交换使主对角线都为1. 3.交换时,可只交换行或只交换列.如:只交换列,行不变(顺序为1,2,3,……,n),那么对于列,只需要根据选择排序,将每行一开始匹配的列的顺序最终也变成1,2,3,……,n即可,因为是选择排序,每次选择第i小的交换到第i个位置…