【poj3279】Fliptile（反转）

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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 × 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".


Input
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
Output
Lines 1.. M: Each line contains N space-separated integers, each specifying how many times to flip that particular location.
Sample Input
4 4
1 0 0 1
0 1 1 0
0 1 1 0
1 0 0 1
Sample Output
0 0 0 0
1 0 0 1
1 0 0 1

0 0 0 0

大意：

输入m行n列的矩阵，让所有的数值都为0，每翻转一个块会让四周跟着翻转，求出每块最少翻转次数的矩阵。

思路：

枚举第一行所有的01情况，共有2^n种，采取从上至下的翻转原则，翻转1下方的块，重复操作直至翻转完最后一行，最后检测最后一行是否全为0，所有满足要求的矩阵输出值最小的。注意不要丢掉IMPOSSIBLE！！！

#include <iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<queue>using namespace std;int n,m,mp[17][17],ansmp[17][17],t[17][17],mp1[17][17];//翻转mp[x][y]和四周的方块void turn(int x,int y){    int nx[5][2]= {1,0,0,1,-1,0,0,-1,0,0};    int i;    for(i=0; i<5; i++)    {        int tx=x+nx[i][0];        int ty=y+nx[i][1];        if(tx>=1&&tx<=m&&ty>=1&&ty<=n)            mp[tx][ty]=!mp[tx][ty];    }}int main(){    cin>>m>>n;    int i,j,c,min1=1e9;    for(i=1; i<=m; i++)        for(j=1; j<=n; j++)            scanf("%d",&mp1[i][j]);    for(c=0; c<=(1<<n); c++)    {        memcpy(mp,mp1,sizeof(mp));//每次循环都要格式化一下mp和t数组        memset(t,0,sizeof(t));//保存翻转次数的矩阵        for(i=1; i<=n; i++)            t[1][i]=c>>(n-i)&1;//枚举第一行所有情况        for(j=1; j<=n; j++)            if(t[1][j]==1)                turn(1,j);//如果是1则翻转mp矩阵        for(i=1; i<m; i++)            for(j=1; j<=n; j++)                if(mp[i][j]==1)                {                    turn(i+1,j);//如果是1翻转下方的块                    t[i+1][j]++;                }        bool flag=0;        for(j=1; j<=n; j++)            if(mp[m][j]==1)                flag=1;        if(flag==1)            continue;        int sum=0;        for(i=1; i<=m; i++)            for(j=1; j<=n; j++)                sum+=t[i][j];        if(min1>sum)        {            min1=sum;            memcpy(ansmp,t,sizeof(ansmp));        }    }    if(min1==1e9)        puts("IMPOSSIBLE");    else    {        for(i=1; i<=m; i++)        {            for(j=1; j<n; j++)                printf("%d ",ansmp[i][j]);            printf("%d\n",ansmp[i][n]);        }    }    return 0;}