Eight Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 30176   Accepted: 13119   Special Judge Description The 15-puzzle has been around for over 100 years; even if you don't know it by that name, you've seen it. It is constructed with 15…

Eight Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 30176   Accepted: 13119   Special Judge Description The 15-puzzle has been around for over 100 years; even if you don't know it by that name, you've seen it. It is constructed with 15…

Eight POJ - 1077 HDU - 1043 八数码问题.用hash(康托展开)判重 bfs(TLE) #include<cstdio> #include<iostream> #include<queue> #include<cstring> using namespace std; ,,,,,,,,,}; ]; ][]; queue<int> q; //data[][9]存储上一个状态,data[][10]存储到这个状态的操作,dat…

先上题目: Eight Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 11243    Accepted Submission(s): 3022Special Judge Problem Description The 15-puzzle has been around for over 100 years; even if you…

http://poj.org/problem?id=1077 http://acm.hdu.edu.cn/showproblem.php?pid=1043 X=a[n]*(n-1)!+a[n-1]*(n-2)!+...+a[i]*(i-1)!+...+a[1]*0! 其中,a[i]为整数,并且X=a[n]*(n-1)!+a[n-1]*(n-2)!+...+a[i]*(i-1)!+...+a[1]*0!.这就是康托展开.康拓展开可以用来表示排列状态,对于本题的9个数字的所有排列只需要9位,所有状态…

http://acm.hdu.edu.cn/showproblem.php?pid=1043 http://poj.org/problem?id=1077 Eight Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 9173    Accepted Submission(s): 2473 Special Judge Problem D…

题目链接:http://poj.org/problem?id=1077 思路分析:题目要求在找出最短的移动路径,使得从给定的状态到达最终状态. <1>搜索算法选择:由于需要找出最短的移动路径,所以选择bfs搜索 <2>判重方法: 将空格视为数字9,则可以将状态的集合视为1-9的排列组合的集合,根据康托展开,将每一个状态映射到一个正整数中: 在由哈希表进行判重. <3>状态压缩:在搜索时使用将每个状态转换为一个由1-9组成的9位数字即可进行状态压缩与解压. <4&g…

题目来源: http://poj.org/problem?id=1077 题目大意: 给你一个由1到8和x组成的3*3矩阵,x每次可以上下左右四个方向交换.求一条路径,得到12345678x这样的矩阵.若没有路径,则输出unsolvable. 经典的八数码问题. 这题我用A*算法做的.推荐一篇博客,从大体上介绍了一下启发式算法的代表A*算法: https://www.cnblogs.com/zhoug2020/p/3468167.html 首先就是判重的问题,搜索的状态是九个数(含x),开个九重…

题目传送门1 2 题意:从无序到有序移动的方案,即最后成1 2 3 4 5 6 7 8 0 分析:八数码经典问题.POJ是一次,HDOJ是多次.因为康托展开还不会,也写不了什么,HDOJ需要从最后的状态逆向搜索,这样才不会超时.判重康托展开,哈希也可. POJ //#include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<string> #include<stack…

Eight Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 18153 Accepted Submission(s): 4908 Special Judge Problem Description The 15-puzzle has been around for over 100 years; even if you don't know…