简单搜索 D题

D - Fliptile

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

题解：

这题也可以dfs+剪枝。
参考了挑战程序设计，很妙的算法。学习了二进制枚举。
详情看代码

代码：

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int MAX_M = 20;const int MAX_N = 20;const int dx[5] = {-1,0,0,0,1};const int dy[5] = {0,-1,0,1,0};//输入int M,N;int tile[MAX_M][MAX_N];int opt[MAX_M][MAX_N];  //保存最优解int flip[MAX_M][MAX_N]; //保存中间结果//查询(x,y)的颜色int get(int x,int y){    int c = tile[x][y];    for(int d=0;d<5;d++)    {        int x2=x+dx[d],y2=y+dy[d];        if(0<=x2&&x2<M&&0<=y2&&y2<N)        {            c += flip[x2][y2];        }    }    return c%2;}//求出第1行确定情况下的最小操作次数//不存在解的话返回-1int calc(){    //求出从第2行开始的翻转方法    for(int i=1;i<M;i++)        for(int j=0;j<N;j++)    {        if(get(i-1,j)!=0)        {            flip[i][j] = 1;        }    }    //判断最后一行是否全白    for(int j=0;j<N;j++)    {        if(get(M-1,j)!=0)        {            return -1;        }    }    //统计翻转的次数    int res=0;    for(int i=0;i<M;i++)        for(int j=0;j<N;j++)    {        res += flip[i][j];    }    return res;}void solve(){    int res=-1;    //按照字典序尝试第一行的所有可能性    for(int i=0;i<(1<<N);i++)    {        memset(flip,0,sizeof(flip));        for(int j=0;j<N;j++)        {            flip[0][N-j-1]=i>>j&1;        }    int num=calc();    if(num>=0&&(res<0||res>num))    {        res = num;        memcpy(opt,flip,sizeof(flip));    }  }    if(res<0)    {        printf("IMPOSSIBLE\n");    }    else    {        for(int i=0;i<M;i++)        {            for(int j=0;j<N;j++)            {                printf("%d%c",opt[i][j],j+1==N?'\n':' ');            }        }    }}int main(){    while(cin>>M>>N)    {        for(int i=0;i<M;i++)            for(int j=0;j<N;j++)            cin>>tile[i][j];        solve();    }    return 0;}