POJ 3041 -- 二分图匹配

题意:给你一个n*m方格 让你用1*2的的小方格去铺满,其中有k个方格不能被铺到. 思路:二分图建图, 以每个格子为点建图,如果可以用一块1*2的小方格铺到,就连一条边. 每个格子在X集合和Y集合都有一个点,只要任意一边被匹配到了就算可以,然后就是二分图匹配了. 上代码. #include<stdio.h> #include<string.h> #include<algorithm> #include<iostream> #include<vector

Asteroids POJ - 3041 Bessie wants to navigate her spaceship through a dangerous asteroid field in the shape of an N x N grid (1 <= N <= 500). The grid contains K asteroids (1 <= K <= 10,000), which are conveniently located at the lattice point

Machine Schedule Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 11922 Accepted: 5077 Description As we all know, machine scheduling is a very classical problem in computer science and has been studied for a very long history. Scheduling p

题意:有两台机器有各有n和m种模式(都是从0开始编号),有k个任务,每个任务可以是第一台机器的x模式或第二台机器的y模式,每台机器切换模式都要重启(重启后初始模式是0),问最少次数重启使所有任务完成. 题解:把任务当做边,两台机器的模式放在不同集合中,那么就是二分图的最小点集覆盖问题,注意模式0可以不计算,因为重启初始模式就是0,不需要花费重启次数. #include <stdio.h> #include <string.h> const int N = 105; int n, m

定理:二分图的最大匹配=最小点覆盖. 思路:将所有行看做点集X,所有列看做点集Y,如果在[i, j]处有小行星,则建立一条从i到j的边,然后求最大匹配即为最小点覆盖数即为答案. 1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 using namespace std; 5 6 const int N = 501; 7 const int M = N * N; 8 int head[N]; 9

题意:给定一个NxN的网格,其中有k个格点上有障碍物,每次可以消除一行或一列障碍物,问最少几次可以消除全部的障碍物. 思路:二分图的经典模型,将所有的行看作二分图中左边的端点,将所有的列看作右边的端点,当格点上有障碍物时,连一条边,问题转变为求二分图的最小顶点覆盖,根据König定理由二分图的最大匹配得到 1 #include<cstdio> 2 #include<iostream> 3 #include<cstring> 4 #include<algorithm

Selecting Courses Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 8667 Accepted: 3863 Description It is well known that it is not easy to select courses in the college, for there is usually conflict among the time of the courses. Li Ming i

Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 17258   Accepted: 9386 Description Bessie wants to navigate her spaceship through a dangerous asteroid field in the shape of an N x N grid (1 <= N <= 500). The grid contains K asteroids (1

题意:图没什么用  给出一个地图 地图上有 点 一次可以覆盖2个连续 的点( 左右 或者 上下表示连续)问最少几条边可以使得每个点都被覆盖 最小路径覆盖       最小路径覆盖=|G|-最大匹配数                   证明:https://blog.csdn.net/qq_34564984/article/details/52778763 证明总的来说就是尽可能多得连边 边越多 可以打包一起处理得点就越多(这里题中打包指连续得两个点只需要一条线段就能覆盖) 拆点思想   :匈牙