POJ 1753 Flip Game【暴搜+深搜】

最新推荐文章于 2020-05-18 20:46:51 发布

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最新推荐文章于 2020-05-18 20:46:51 发布

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原文链接：http://www.cnblogs.com/dream-wind/archive/2012/03/15/2397836.html

本文介绍了一款4x4棋盘游戏，玩家通过选择一个方块并翻转其相邻方块的颜色来实现所有方块颜色一致的目标。文章提供了两种求解方法：二进制法和深度优先搜索，并详细阐述了实现过程。

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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

二进制法：

code：

View Code

#include<stdio.h>
#define max 999999
char s[4][4];
int cs[16]={19,39,78,140,305,626,1252,2248,4880,8992,20032,35968,12544,29184,58368,51200};
int po[16] = {1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768};
int main()
{
int i,j,value=0,tot,value1;
int min=max;
char c;
for(i=0;i<4;i++)
    {
        scanf("%s",s[i]);
for(j=0;j<4;j++)
if(s[i][j]=='b')
                value+=po[4*i+j];
else continue;
    }
for(i=0;i<65535;i++)
    {
        tot=0;
        value1=value;
for(j=0;j<16;j++)
if(i&po[j])
            {
                tot++;
                value1=value1^cs[j];
            }
if(value1==0||value1==65535)
                min=min<tot?min:tot;
    }
if(min==max)
        printf("Impossible\n");
else
        printf("%d\n",min);
return 0;
}

深搜：
code：

View Code

#include<stdio.h>
char map[4][5];
int s[4][4];
int f[10]={-1,0,0,0,1,0,-1,0,1,0};
int tot=20;
void dfs(int p,int c)
{
int i,j,k=1;
for(i=0;k&&i<4;i++)
for(j=0;k&&j<4;j++)
if(s[i][j]!=s[0][0])
                k=0;
if(k)
    {
if(c<tot)
            tot=c;
return;
    }
if(p>15)
return;
    dfs(p+1,c);
for(k=0;k<5;k++)
    {
        i=p/4+f[k];
        j=p%4+f[k+5];
if(i>=0&&i<4&&j>=0&&j<4)
            s[i][j]=1-s[i][j];
    }
    dfs(p+1,c+1);
for(k=0;k<5;k++)
    {
        i=p/4+f[k];
        j=p%4+f[k+5];
if(i>=0&&i<4&&j>=0&&j<4)
            s[i][j]=1-s[i][j];
    }
}
int main()
{
int i,j;
for(i=0;i<4;i++)
    {
        scanf("%s",map[i]);
for(j=0;j<4;j++)
if(map[i][j]=='b')
                s[i][j]=1;
else
                s[i][j]=0;
    }
    dfs(0,0);
if(tot<20)
        printf("%d\n",tot);
else
        printf("Impossible");
return 0;
}

转载于:https://www.cnblogs.com/dream-wind/archive/2012/03/15/2397836.html