Consider the following position as an example: bwbw wwww bbwb bwwb Here "b" denotes pieces lying their black side up and "w" denotes pieces lying their white side up. If we choose to flip the 1st piece from the 3rd row (this choice is shown at the picture), then the field will become: bwbw bwww wwwb wwwb The goal of the game is to flip either all pieces white side up or all pieces black side up. You are to write a PRogram that will search for the minimum number of rounds needed to achieve this goal. InputThe input consists of 4 lines with 4 characters "w" or "b" each that denote game field position.OutputWrite to the output file a single integer number - the minimum number of rounds needed to achieve the goal of the game from the given position. If the goal is initially achieved, then write 0. If it's impossible to achieve the goal, then write the Word "Impossible" (without quotes).Sample InputbwwbbbwbbwwbbwwwSample Output4解题报告
不要想太复杂就好，要明确如果有答案，那么答案小于等于8#include<stdio.h>#include<string.h>bool map[6][6];int ox[]={0,0,0,1,-1};int oy[]={0,1,-1,0,0};int best;bool check(){ //end check for(int j=1;j<=4;j++) for(int k=1;k<=4;k++) if(map[1][1]!=map[j][k]) return false; return true;}void op(int X,int Y){ for(int i=0;i<5;i++){ int x=X+ox[i]; int y=Y+oy[i]; map[x][y]=!map[x][y]; }}void dfs(int t,int cnt){ if(check()){ if(cnt<best) best=cnt; return ; } if(cnt>best||t>16) return ; int X=t%4+1; int Y=t/4+1; //剪枝 加上该条件 78ms立马变0ms if(Y>1&&X>1&&map[X-1][Y-1]!=map[1][1]) return ; dfs(t+1,cnt); op(X,Y); dfs(t+1,cnt+1); op(X,Y);}int main(){ char str[5]; for(int i=1;i<=4;i++){ scanf("%s",str); for(int j=1;j<=4;j++) map[i][j]=str[j-1]=='b'; } best=9; dfs(0,0); if(best<9) printf("%d/n",best); else puts("Impossible"); return 0;}
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