HDU(1753)

Flip Game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 27202   Accepted: 11803 Description Flip game is played on a rectangular 4x4 field with two-sided pieces placed on each of its 16 squares. One side of each piece is white and the other one is black and each piece is lying either it's black or white side up. Each round you flip 3 to 5 pieces, thus changing the color of their upper side from black to white and vice versa. The pieces to be flipped are chosen every round according to the following rules:  Choose any one of the 16 pieces.  Flip the chosen piece and also all adjacent pieces to the left, to the right, to the top, and to the bottom of the chosen piece (if there are any). 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.  Input The input consists of 4 lines with 4 characters "w" or "b" each that denote game field position. Output Write 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 Input bwwb bbwb bwwb bwww Sample Output 4 题目大意：一个硬币有正、反两个面，在一个  4x4 的方格中 ， 放着16枚硬币，每枚硬币是正面还是反面是随机的，当任意翻转其中一枚硬币时，这枚硬币的上、下、左、右都要翻转。问至少翻转几枚硬币可以使16枚硬币都是正面或都是反面。 #include <iostream> #include <cstdio> using namespace std; int ar[6][6]={0}; bool flag; void Input(){ //输入初始棋盘 char c; for(int i=1;i<5;i++){ for(int j=1;j<5;j++){ scanf("%c",&c); if(c=='w') ar[i][j]=1; } getchar(); } } bool judge(){ //判断是否全为黑 or 白 for(int i=1;i<5;i++){ for(int j=1;j<5;j++){ if(ar[1][1]!=ar[i][j]) return false; } } return true; } void filp(int r,int c){//翻旗 ar[r][c]=!ar[r][c]; ar[r+1][c]=!ar[r+1][c]; ar[r][c+1]=!ar[r][c+1]; ar[r-1][c]=!ar[r-1][c]; ar[r][c-1]=!ar[r][c-1]; } void dfs(int r,int c,int time,int max){ if(flag)return; //若已成功，直接返回 if(time == max || c==5){ //当走完规定步数、或全部翻完时结束 flag=judge(); return; } //翻当前棋的情况 filp(r,c); if(r<4){//先按行枚举、再按列枚举 dfs(r+1,c,time+1,max); } else{ dfs(1,c+1,time+1,max); } //不翻当前棋的情况 filp(r,c); if(r<4){ dfs(r+1,c,time,max); } else{ dfs(1,c+1,time,max); } } int main() { Input(); flag=false; for(int i=0;i<17;i++){ dfs(1,1,0,i); if(flag){ printf("%d\n",i); break; } } if(!flag)printf("Impossible\n"); return 0; } 在这个程序中，有很多地方需要注意，首先在 Input 函数中 ，getchar()用于提取输入中无用的字符；可以在主函数外申请全局的数组。最灵活的是这步： if(r<4 && c<4) dfs(r+1,c,time+1,max); else dfs(1,c+1,time+1,max); 这一步把16枚硬币一个一个的遍历了一遍。 这个程序通过枚举各种情况，最终找到答案。效果很好，很值得借鉴。