题目网址:http://poj.org/problem?id=1753 题目: Flip Game Description Flip game is played on a rectangular 4x4 field with two-sided pieces placed on each of its 16 squares. One side of each piece is white and the other one is black and each piece is lying ei…

Flip game is played on a rectangular 4x4 field with two-sided pieces placed on each of its 16 squares. One side of each piece is white and the other one is black and each piece is lying either it's black or white side up. Each round you flip 3 to 5 p…

Flip Game Time Limit: 1000MS  Memory Limit: 65536K  Total Submissions: 4863  Accepted: 1983 Description Flip game is played on a rectangular 4x4 field with two-sided pieces placed on each of its 16 squares. One side of each piece is white and the oth…

题目链接 有一个n*m(1<=n,m<=20)的网格图,图中有k堵墙和有一条长度为L(L<=8)的蛇,蛇在移动的过程中不能碰到自己的身体.求蛇移动到点(1,1)所需的最小步数. 显然用8个(x,y)来表示蛇的状态是不现实的(用哈希也很难存下,要么爆内存,要么超时),所以首先应当进行状态压缩.可以发现蛇的身体是连续的,因此可以用一个表示方向的向量来储存蛇身体的每个部分在它上个部分的哪个方向(只有头部用(x,y)表示),这样总状态数就变成了n*m*(2^((L-1)*2)),在可接受范围内了…

题意:有n座城市和m(1<=n,m<=10)条路.现在要从城市1到城市n.有些路是要收费的,从a城市到b城市,如果之前到过c城市,那么只要付P的钱, 如果没有去过就付R的钱.求的是最少要花多少钱. 析:BFS,然后由于走的路线不同,甚至边或者点都可能多走,所以用状态压缩.然后本题有坑啊,有重连,而且有很多条重边,所以多走几次就好了. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include…

Count Color Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 38921   Accepted: 11696 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.…

题目:http://poj.org/problem?id=1753 因为粗心错了好多次……,尤其是把1<<15当成了65535: 参考博客:http://www.cnblogs.com/kuangbin/archive/2011/07/30/2121677.html 主要思想: 1.如果用一个4*4的数组存储每一种状态,不但存储空间很大,而且在穷举状态时也不方便记录.因为每一颗棋子都只有两种状态,所以可以用二进制0和1表示每一个棋子的状态,则棋盘的状态就可以用一个16位的整数唯一标识.而翻转的…

题目地址:http://poj.org/problem?id=1753 /* 这题几乎和POJ 2965一样,DFS函数都不用修改 只要修改一下change规则... 注意:是否初始已经ok了要先判断 */ #include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <cmath> #include <string> #i…

胜利大逃亡(续) Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 7357    Accepted Submission(s): 2552 Problem Description Ignatius再次被魔王抓走了(搞不懂他咋这么讨魔王喜欢)……这次魔王汲取了上次的教训,把Ignatius关在一个n*m的地牢里,并在地牢的某些地方安装了带…

1.题意:一个由01组成的4*4的矩阵,可以实现相邻元素交换位置的操作,给出初试状态和目标状态,试求最少操作数的方案: 2.输入输出:输入给出初试矩阵和目标矩阵:要求输出最小操作的次数: 3.分析:输出最小操作数,很容易联想到使用BFS,这里为了方便表示,把4*4的矩阵拉成一个16个数的数组来看,并用一个16位二进制数表示其状态:用位运算来实现交换某两位的状态,另外再稍微注意一下如何在表示"相邻"的概念即可: # include <iostream> # include &…