poj Flip Game 1753 (枚举)】的更多相关文章

Flip Game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 27005   Accepted: 11694 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…
看题传送门:http://poj.org/problem?id=1753 DFS枚举的应用. 基本上是参考大神的.... 学习学习.. #include<cstdio> #include<iostream> using namespace std; int n,s,d,ans; bool a[5][5],flag=false; //判断全部一样的情况 bool alllike() { for(int i=0;i<4;i++) for(int j=0;j<4;j++) i…
题目:http://poj.org/problem?id=1753 这个题在開始接触的训练计划的时候做过,当时用的是DFS遍历,其机制就是把每一个棋子翻一遍.然后顺利的过了.所以也就没有深究. 省赛前一次做PC2遇到了差点儿一模一样的题,仅仅只是是把棋盘的界限由4X4改为了5X5,然后一直跑不出结果来,可是当时崔老湿那个队过了,在最后总结的时候.崔老湿就说和这个题一样,只是要枚举第一行进行优化. 我以为就是恢复第一行然后第二行以此类推,只是手推一下结果是6不是4,就知道这个有问题. 问了崔老湿,…
题目链接: http://poj.org/problem?id=1753 题意: 由白块黑块组成的4*4方格,每次换一个块的颜色,其上下左右的块也会被换成相反的颜色.问最少换多少块,使得最终方格变为全白或者全黑~ 分析: 典型的枚举. 这种问题被称为开关问题,关键是要对周围的块翻动对自身的影响进行记录.后一排决定了前一排的最终状态,枚举第一排的情况即可. 代码: #include<iostream> #include<cstring> using namespace std; co…
Flip Game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 26492   Accepted: 11422 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…
Flip Game   Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 52279   Accepted: 22018 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 th…
Flip Game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 40632   Accepted: 17647 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…
题目链接:http://poj.org/problem?id=2784 Buy or Build Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 1528   Accepted: 592 Description World Wide Networks (WWN) is a leading company that operates large telecommunication networks. WWN would li…
http://poj.org/problem?id=3279 题意 一个m*n的01矩阵,每次翻转(x,y),那么它上下左右以及本身就会0变1,1变0,问把矩阵变成全0的,最小需要点击多少步,并输出最小字典序的操作. 分析 爆搜铁定超时...对于一个格子,要么反转一次,要么反转零次.反转的顺序不改变最终结果.那么我们试着枚举第一行的反转情况(状态压缩),此时能影响第一行的只有第二行的格子了,依次类推,一行一行来.最后检测最后一行的值,就知道了此状态可行不可行.枚举时可以确定反转的字典序从小到大.…
题目链接:http://poj.org/problem?id=2912 Time Limit: 5000MS Memory Limit: 65536K Description N children are playing Rochambeau (scissors-rock-cloth) game with you. One of them is the judge. The rest children are divided into three groups (it is possible t…