Fliptile 翻格子游戏[Usaco2007 Open]

题目描述

Farmer John knows that an intellectually satisfied cow is a happy cow who will give more milk. He has arranged a brainy activity for cows in which they manipulate an M x N grid (1 <= M <= 15; 1 <= N <= 15) of square tiles, each of which is colored black on one side and white on the other side. As one would guess, when a single white tile is flipped, it changes to black; when a single black tile is flipped, it changes to white. The cows are rewarded when they flip the tiles so that each tile has the white side face up. However, the cows have rather large hooves and when they try to flip a certain tile, they also flip all the adjacent tiles (tiles that share a full edge with the flipped tile). Since the flips are tiring, the cows want to minimize the number of flips they have to make. Help the cows determine the minimum number of flips required, and the locations to flip to achieve that minimum. If there are multiple ways to achieve the task with the minimum amount of flips, return the one with the least lexicographical ordering in the output when considered as a string. If the task is impossible, print one line with the word "IMPOSSIBLE".

约翰知道，那些高智力又快乐的奶牛产奶量特别高．所以他做了一个翻瓦片的益智游戏来娱乐奶牛．在一个M×N(1≤M，N≤15)的骨架上，每一个格子里都有一个可以翻转的瓦片．瓦片的一面是黑色的，而另一面是白色的．对一个瓦片进行翻转，可以使黑变白，也可以使白变黑．然而，奶牛们的蹄子是如此的巨大而且笨拙，所以她们翻转一个瓦片的时候，与之有公共边的相邻瓦片也都被翻转了．那么，这些奶牛们最少需要多少次翻转，使所有的瓦片都变成白面向上呢？如杲可以做到，输出字典序最小的结果（将结果当成字符串处理）．如果不能做到，输出“IMPOSSIBLE”．

输入

* Line 1: Two space-separated integers: M and N

* Lines 2..M+1: Line i+1 describes the colors (left to right) of row i of the grid with N space-separated integers which are 1 for black and 0 for white

第1行输入M和N，之后M行N列，输入游戏开始时的瓦片状态．0表示白面向上，1表示黑面向上．

输出

* Lines 1..M: Each line contains N space-separated integers, each specifying how many times to flip that particular location.

输出M行，每行N个用空格隔开的整数，表示对应的格子进行了多少次翻转．

样例输入

样例输出

0 0 0 0
1 0 0 1
1 0 0 1
0 0 0 0OUTPUT DETAILS:
After flipping at row 2 column 1, the board will look like:
0 0 0 1
1 0 1 0
1 1 1 0
1 0 0 1
After flipping at row 2 column 4, the board will look like:
0 0 0 0
1 0 0 1
1 1 1 1
1 0 0 1
After flipping at row 3 column 1, the board will look like:
0 0 0 0
0 0 0 1
0 0 1 1
0 0 0 1After flipping at row 3 column 4, the board will look like:
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0Another solution might be:
0 1 1 0
0 0 0 0
0 0 0 0
0 1 1 0
but this solution is lexicographically higher than the solution above.

题解一眼以为是喜闻乐见的每日大暴搜，然后就开心地开始画。这个东西和翻的先后无关，而且每个点翻两次就翻回去了，只会有0和1两种情况。直接按字典序从小到大翻不翻枚举，得到第一个可行解就停止搜索，没有想清楚“步数最少”到底是不是题目要求。稍微剪了剪枝，就直接这么放着了。搜索复杂度比较玄学，也没仔细想，天真地以为这样就差不多了。    所以说我这么信任你你居然是个状压？用二进制枚举第一行的操作，正如《分手是祝愿》一样，每个点被翻偶数次状态与原来相同。用一个矩阵记录一下每个位置被改变了多少次，在翻下一行时要求上一行全部为0，而能改变的上一行只有正对的那个，所以下一行的操作也是可以确定的。推到最后一行，这一系列操作真正正确与否就可以验证了，看它是否使最后一行全部变为0。当搜出了一个可行解，它固然字典序很小，但还要比较操作次数，并且以操作次数小的优先(话说字典序大多数情况下都是用来优中选优的)，所以无论如何都要在第一行枚举所有状态。    昨天找规律，今天递推，正解还真是五花八门多种多样丰富多彩惊世骇俗啊……然而真正的dalao们还是能想到解决问题的办法。存在即合理，有题目就会有解法，有偏差就会有错误，我们总是得相信些什么，有所相信才能走下去。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int ans[18],a[18][18],n,m,bf[18][18],res[18],mi,ge;
bool jg,cl;
void init()
{scanf("%d%d",&m,&n);for(int i=1;i<=m;i++)for(int j=1;j<=n;j++)scanf("%d",&a[i][j]);mi=m*n+1;
}
void dt()
{for(int j=n-1;j>=0;j--)if(ans[1]&(1<<j)) {bf[1][n-j]++;bf[1][n-j-1]++;bf[1][n-j+1]++;bf[2][n-j]++;}for(int i=2;i<=m;i++)for(int j=1;j<=n;j++)if((bf[i-1][j]&1)!=a[i-1][j]){ans[i]|=(1<<(n-j));bf[i][j]++;bf[i][j+1]++;bf[i][j-1]++;bf[i+1][j]++;}cl=1;for(int j=1;j<=n;j++)if((bf[m][j]&1)!=a[m][j]){  cl=0;break;}
}
int main()
{init();for(int i=0;i<=(1<<n)-1;i++){memset(bf,0,sizeof(bf));memset(ans,0,sizeof(ans));ans[1]=i;dt();if(cl){jg=1;ge=0;for(int k=1;k<=m;k++)for(int j=0;j<n;j++)if(ans[k]&(1<<j)) ge++;if(ge<mi){for(int k=1;k<=m;k++)res[k]=ans[k];mi=ge;}} }if(jg)for(int i=1;i<=m;i++){for(int j=n-1;j>0;j--){if(res[i]&(1<<j))  printf("1 ");else  printf("0 ");}if(res[i]&1) printf("1\n");else  printf("0\n");}if(!jg) printf("IMPOSSIBLE");return 0;
}

fliptile

食堂知道，那些疲惫又难过的OIer吃得特别多，所以他们很喜欢大暴搜……

转载于:https://www.cnblogs.com/moyiii-/p/7275262.html