开始时候只用了BFS,显然超时啊,必然在结构体里加一个数组什么的判重啊,开始用的一个BOOL数组,显然还是不行,复杂度高,每次都要遍历数组来判重:后百度之,学习了二进制状态压缩,其实就用一个二进制数来表示当前状态,然后就可以用简单快速位运算来操作了. #include<map> #include<iostream> #include<queue> #include<cstring> #include<cstdio> using namespace…

好久没写搜索题了,就当练手吧. vis[][][1025]第三个维度用来维护不同key持有状态的访问情况. 对于只有钥匙没有对应门的位置,置为'.',避免不必要的状态分支. // // main.cpp // hdu_1429 // // Created by Luke on 16/10/8. // Copyright © 2016年 Luke. All rights reserved. // #include <iostream> #include <string> #inclu…

Count Color Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 53312   Accepted: 16050 Description Chosen Problem Solving and Program design as an optional course, you are required to solve all kinds of problems. Here, we get a new problem.…

最短Hamilton路径(二进制状态压缩) 题目描述:n个点的带权无向图,从0-n-1,求从起点0到终点n-1的最短Hamilton路径(Hamilton路径:从0-n-1不重不漏的每个点恰好进过一次) 题解:二进制状态压缩算法\(O(2^n*n^2)\),需要记录当前经过了哪些点,当前在哪个位置.\(f[i][j]\)   \(i\)转化为二进制每一位代表是否经过该点,\(j\)表示当前位于j这个点 #include <iostream> #include <cstring> u…

Problem Description Rompire is a robot kingdom and a lot of robots live there peacefully. But one day, the king of Rompire was captured by human beings. His thinking circuit was changed by human and thus became a tyrant. All those who are against him…

题目链接 864. 获取所有钥匙的最短路径 题意 给定起点,要求在最短步骤内收集完所有钥匙,遇到每把锁之前只有 有对应的钥匙才能够打开 思路 BFS+状态压缩典型题目 先确定起点和总的钥匙数目,其次难点有两处: 如何确定当前路径下已经收集好特定的钥匙 如何确定钥匙已经全部收集完成 第一个问题:可以把每一个节点的状态定义为(x,y,state),其中state为钥匙数目的二进制表示,例如现在收集'b'这把钥匙,那么state更新为2(000010),括号里面为钥匙的二进制数,如果下一轮遇到'B',…

Description Several startup companies have decided to build a better Internet, called the "FiberNet". They have already installed many nodes that act as routers all around the world. Unfortunately, they started to quarrel about the connecting li…

http://poj.org/problem?id=3311 Hie with the Pie Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 4456   Accepted: 2355 Description The Pizazz Pizzeria prides itself in delivering pizzas to its customers as fast as possible. Unfortunately,…

Swipe Bo Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1549    Accepted Submission(s): 315 Problem Description "Swipe Bo" is a puzzle game that requires foresight and skill.  The main ch…

昨天晚上12点刷到的这个题,一开始一位是BFS,但是一直没有思路.后来推了一下发现只需要依次枚举第一行的所有翻转状态然后再对每个情况的其它田地翻转进行暴力dfs就可以,但是由于二进制压缩学的不是很透,一直有小问题,下面我还会讲子集生成的相关方法,有兴趣的同学可以继续关注. 本题大意:一块地,有黑(1)白(0)之分,牛每次踩踏使得当前块以及四连块都变色,问当牛如何踩时使得地都变白并且求出最小踩踏次数和踩踏路径的最小字典序时的踩踏地图. 本题思路:由于同一块地被翻两次都会回到原来的状态,所以只需要对…