最大匹配正好是满足连接关系的最大数目,所以用匈牙利算法即可求 二分图代码: #include <cstdio>#include<iostream>#include<algorithm>#include<cstring>usingnamespacestd;charG[5][5];intmap[20][20],col[5][5],row[5][5];intmatch[20],vis[20];intcntx ,cnty,n;boolFind(intu){for(inti=0; i<...
思路: 由于n不大于4,最多16个点,想到可以二进制枚举子集,然后逐个判断每个子集的可行性。 1#include<iostream>2#include<string>3#include<vector>4#include<cstdio>5#include<cstring>6#defineFOR(i, a, b) for(int i = a; i < b; i++)7usingnamespacestd;8intn, k;9chara[10][10];10intdir[...
问最多能放置多少个碉堡,使它们彼此不会互相摧毁。 Problem solving:从起点(0,0)开始一直到(n-1,n-1)遍历每一个点,每个都有放与不放两种情况,并且也要判断每个点是否可以放碉堡。当判断是否可以放的时候,可判断此行此列以前遍历的时候是否放过,往左、上方判断,不必往下、右方判断,因为下面和右边还未遍历。
HDU - 1045 Fire Net (缩点建图+二分图) Fire Net Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 14784 Accepted Submission(s): 8936 Problem Description Suppose that we have a square city with straight streets. A map of a city is...
【HDU - 1045】Fire Net (dfs 或二分图) 题干: 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...
HDU_1045 Fire Net Fire Net Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 18035 Accepted Submission(s): 10996 题意: 一个二维城市中有街道和墙,子弹可以穿过街道,但是会完全被墙住,要在其中放入尽量多的炮台,炮台向正交方向发射子弹,问要使...
【多帧超分辨率(MFSR)网络HighRes-net】’HighRes-net - Pytorch implementation of HighRes-net, a neural network for multi frame super-resolution (MFSR), trained and tested on the European Space Agency’s Kelvin competition.' by ElementAI GitHub: O网页链接 ...
HDU 1045 Fire Net 题目传送门:https://vjudge.net/problem/HDU-1045 题意:在一个n*n(n<=4)的区域内建造炮塔,炮塔与炮塔不能在同一行和同一列除非中间有城墙阻隔,要你求出最多能建造多少个炮塔。 思路:因为n最大只有4,可以利用dfs全面检索。......
char map[5][5]; int n,ans; bool leagal(int x,int y) { bool have; have = false; map[x][y] = 'B'; for(int i=0;i<n;i++) { if(map[x][i] == 'B') if(have) return false; else have = true; if(map[x][i] == 'X') ...
HDU - 1045 Fire Net 就是转换成,求最大匹配,附代码 /*** 二分图匹配(匈牙利算法的DFS实现) INIT:g[][]两边定点划分的情况 CALL:res=hungary();输出最大匹配数优点:适于稠密图,DFS找增广路快,实现简洁易于理解时间复杂度:O(VE); ***/ #include<iostream> using namespace std; #include<cstdio> #...