poj.1753dfs

这道题的意思很简单，主要需要注意的有两点，第一：矩阵中的每个点都最多被主动翻一次，（注意不是被动），所以总共有2^16种可能，只要枚举每种可能了。第二：翻转的顺序对结果没有影响，所以问题的关键就在于那些点被主动翻转了。翻转多少次就有多少次round,只要求出所有round中最小的一个就是答案了，如果没有一种可能可以得到结果则输出“Impossible"。下面是代码和测试数据：

//#include <iostream>#include <stdio.h>#include <stdlib.h>//using namespace std;#define Inf 1000000int maxn;int s[4][4];void Init(){char ch[4][4];int i,j;for(i=0;i<4;i++)scanf("%s",ch[i]);for(i=0;i<4;i++)for(j=0;j<4;j++){if(ch[i][j]=='b')s[i][j]=1;elses[i][j]=-1;}/*for(i=0;i<4;i++)for(j=0;j<4;j++)             printf("%d ",s[i][j]);*/}bool Is_right(int (*t)[4]){int i,j;for(i=0;i<4;i++)for(j=0;j<4;j++)if(t[i][j]!=t[0][0])return false;return true;}void dfs(int (*t)[4],int x,int y,int deep){if(Is_right(t)){if(deep<maxn)maxn=deep;return ;}if(x>=4)return ;//int (*record)[4]=t;int record[4][4],i,j;for(i=0;i<4;i++)for(j=0;j<4;j++)record[i][j]=t[i][j];if(x>=0 && x<=3 && y>=0 && y<=3)record[x][y]=-record[x][y];if(x-1>=0 && x-1<=3 && y>=0 && y<=3)record[x-1][y]=-record[x-1][y];if(x+1>=0 && x+1<=3 && y>=0 && y<=3)record[x+1][y]=-record[x+1][y];if(x>=0 && x<=3 && y-1>=0 && y-1<=3)record[x][y-1]=-record[x][y-1];if(x>=0 && x<=3 && y+1>=0 && y+1<=3)record[x][y+1]=-record[x][y+1];if(y<=2){dfs(t,x,y+1,deep);dfs(record,x,y+1,deep+1);}else{dfs(t,x+1,0,deep);dfs(record,x+1,0,deep+1);}}int main(){//int i,j;while(true){     Init();    if(Is_right(s)){printf("0\n");continue;//return 0;}maxn=Inf;dfs(s,0,0,0);if(maxn==Inf){printf("Impossible\n");continue;//return 0;}printf("%d\n",maxn);}return 0;}

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