题目地址:http://poj.org/problem?id=2965 /* 题意:4*4的矩形,改变任意点,把所有'+'变成'-',,每一次同行同列的都会反转,求最小步数,并打印方案 DFS:把'+'记为1, '-'记为0 1. 从(1, 1)走到(4, 4),每一点DFS两次(改点反转或不反转):used记录反转的位置 详细解释:http://poj.org/showmessage?message_id=346723 2. 比较巧妙的解法:抓住'+'位置反转,'-'位置相对不反转的特点,从状…

The Pilots Brothers' refrigerator Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 22286   Accepted: 8603   Special Judge Description The game “The Pilots Brothers: following the stripy elephant” has a quest where a player needs to open a…

The Pilots Brothers' refrigerator Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 16868   Accepted: 6393   Special Judge Description The game “The Pilots Brothers: following the stripy elephant” has a quest where a player needs to open a…

  The Pilots Brothers' refrigerator Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 15176   Accepted: 5672   Special Judge Description The game “The Pilots Brothers: following the stripy elephant” has a quest where a player needs to open…

The Pilots Brothers' refrigerator Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 15136   Accepted: 5660   Special Judge Description The game “The Pilots Brothers: following the stripy elephant” has a quest where a player needs to open a…

The Pilots Brothers' refrigerator Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 17450   Accepted: 6600   Special Judge Description The game “The Pilots Brothers: following the stripy elephant” has a quest where a player needs to open a…

The game “The Pilots Brothers: following the stripy elephant” has a quest where a player needs to open a refrigerator. There are 16 handles on the refrigerator door. Every handle can be in one of two states: open or closed. The refrigerator is open o…

题目链接: http://poj.org/problem?id=1753 题意: 给定冰箱门的开关情况,改变一个门则其所在行列的门都会发生改变,求出改变门的最少操作使得最终所有门都是打开状态. 代码: bfs+状态压缩很容易想到~~ 这里的状态压缩要需要多加小心,注意一下存储的是翻转门的情况~ #include<iostream> #include<cstdio> #include<queue> using namespace std; typedef pair<…

//题目:http://poj.org/problem?id=2965//题意:电冰箱有16个把手,每个把手两种状态(开‘-’或关‘+’),只有在所有把手都打开时,门才开,输入数据是个4*4的矩阵,因此考虑用位表示.可以改变任意一个把手的位置,但同时改变其所在的行和列.求最小步骤.//耗时 800MS1 #include<stdio.h> #include<string.h> #include<stdlib.h> #include<queue> using…

http://poj.org/problem?id=2965 题意: 一个4*4的矩形,有'+'和'-'两种符号,每次可以转换一个坐标的符号,同时该列和该行上的其他符号也要随之改变.最少需要几次才能全部变成'-'. 思路: 这道题和黑白棋那道题目差不多,唯一的差别就是需要记录一下路径. 我是用BFS来做的,用二进制来存储.翻转时用位运算的异或比较方便,可以看一下我的这篇博客(http://www.cnblogs.com/zyb993963526/p/6347741.html),上面写的比较清楚.…