hdu 1195 Open the Lock

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题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=1195 Open the Lock Description Now an emergent task for you is to open a password lock. The password is consisted of four digits. Each digit is numbered from 1 to 9. Each time, you can add or minus 1 to…

hdu - 1195 Open the Lock (bfs) && hdu 1973 Prime Path (bfs)

http://acm.hdu.edu.cn/showproblem.php?pid=1195 这道题虽然只是从四个数到四个数,但是状态很多,开始一直不知道怎么下手,关键就是如何划分这些状态,确保每一个状态都能遍历到. 得到四个数之后,分三种情况处理,每次改变一个数之后都要加入队列,最先输出的就是步数最少. #include <cstdio> #include <cstring> #include <queue> using namespace std; struct p…

hdu 1195:Open the Lock（暴力BFS广搜）

Open the Lock Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3847 Accepted Submission(s): 1661 Problem Description Now an emergent task for you is to open a password lock. The password is co…

hdu 1195 Open the Lock (BFS)

Open the Lock Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3388 Accepted Submission(s): 1499 Problem Description Now an emergent task for you is to open a password lock. The password is co…

hdu 1195 Open the Lock（广搜，简单）

题目猜密码,问最少操作多少次猜对,思路很简单的广搜,各种可能一个个列出来就可以了,可惜我写的很搓. 不过还是很开心,今天第一个一次过了的代码 #define _CRT_SECURE_NO_WARNINGS //这是非一般的最短路,所以广搜到的最短的路不一定是所要的路线 //所以应该把所有的路径都搜索出来,找到最短的转折数,看他是不是不大于2 //我是用边搜索边更新当前路径的最小转弯数来写的 #include<stdio.h> #include<string.h> #inclu…

HDU 1195 Open the Lock （双宽搜索）

意甲冠军:给你一个初始4数字和目标4数字,当被问及最初的目标转换为数字后,. 变换规则:每一个数字能够加1(9+1=1)或减1(1-1=9),或交换相邻的数字(最左和最右不是相邻的). 双向广搜:分别对初始和目标数字进行广搜,vis数组用1和2标记两种已搜索的数字,用mp数组记录状态的步数. 当从前往后搜能够到达2或从后往前搜能够到达1状态则就能够了... #include<stdio.h> #include<string.h> #include<string> #in…

[HUD 1195] Open the Lock

Open the Lock Problem Description Now an emergent task for you is to open a password lock. The password is consisted of four digits. Each digit is numbered from 1 to 9. Each time, you can add or minus 1 to any digit. When add 1 to '9', the digit will…