Flip Game POJ
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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:
1.Choose any one of the 16 pieces.
2.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
题意:翻转棋子,翻任意一个棋子时,这枚棋子的上下左右也要翻转,如果可以翻出全部白色或者黑色,输出最小的步数(可能是0),不能的话输出impossible
#include <stdio.h>#include <string.h>#include <stdlib.h>#include <iostream>using namespace std;#define inf 9999999int map[10][10];int min_dis=inf;void change(int x,int y)//翻转棋子{ map[x][y]=1-map[x][y]; map[x+1][y]=1-map[x+1][y]; map[x][y+1]=1-map[x][y+1]; //数组开大一点 节约时间 if(x>0) map[x-1][y]=1-map[x-1][y]; if(y>0) map[x][y-1]=1-map[x][y-1];}int check()//查看是否是同一颜色{ int i,j; int cnt=0; for(i=0;i<4;i++) for(j=0;j<4;j++) cnt+=map[i][j]; if(cnt==0||cnt==16)//棋盘为全白或全黑 return 1; else return 0;}void search(int d,int step){ int x,y; if(d==16) { if(check() && step<min_dis) min_dis=step; } else{ x=d/4; y=d%4; search(d+1,step); //判断 不用再翻别的棋子 就满足要求的情况 change(x,y); search(d+1,step+1); change(x,y); } return;}int main(){ char str; int i,j; for(i=0;i<4;i++) for(j=0;j<4;j++) { cin>>str; if(str=='b') map[i][j]=1; else map[i][j]=0; } search(0,0); if(min_dis == inf) printf("Impossible\n"); else printf("%d\n",min_dis); return 0;}
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