POJ1753——DFS+枚举——Flip Game

最新推荐文章于 2020-10-20 22:40:40 发布

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最新推荐文章于 2020-10-20 22:40:40 发布

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原文链接：http://www.cnblogs.com/zero-begin/p/4313118.html

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本文深入解析了一种名为Flipgame的游戏算法，该游戏在一个4x4的棋盘上进行，玩家的目标是通过翻转棋子，使所有棋子的颜色统一。文章详细介绍了游戏规则，包括如何选择并翻转棋子，以及如何计算最少的翻转次数来达到游戏目标。通过使用C++编程语言，文章提供了一个解决此问题的算法实现，包括状态判断和深度优先搜索策略。

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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

Source

Northeastern Europe 2000

大意：翻转一个棋子，周围的棋子也要改变，求最小翻的次数

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int a[6][6];
int step,flag;
int judge(int a[6][6]){
    for(int i = 1; i <= 4; i++){
            for(int j = 1; j <= 4; j ++){
                    if(a[i][j] != a[1][1])
                     return 0;
            }
    }
    return 1;
}
void fanzhuan(int i,int j){
    a[i][j] = !a[i][j];
    a[i][j+1] = !a[i][j+1];
    a[i][j-1] = !a[i][j-1];
    a[i+1][j] = !a[i+1][j];
    a[i-1][j] = !a[i-1][j];
}
void dfs(int i, int j,int foot){
    if(foot == step)
    {
        flag = judge(a);
        if(flag == 1) return ;
    }
    if(flag||i == 5) return ;
    fanzhuan(i,j);
    if(j<4)
    dfs(i,j+1,foot+1);
    else dfs(i+1,1,foot+1);
    fanzhuan(i,j);
    if(j<4)
    dfs(i,j+1,foot);
    else dfs(i+1,1,foot);
    return ;
}
int main(){
    char t;
    int i,j;
    memset(a,0,sizeof(a));
    for( i = 1; i <= 4 ; i++){
        for( j = 1; j <= 4; j++){
              scanf("%c",&t);
              if(t == 'b') a[i][j] = 1;
        }
        getchar();
    }
    for(step = 0; step<= 16; step++){
        dfs(1,1,0);
        if(flag==1)
            break;
    }
    if(step>16)
        printf("Impossible\n");
    else printf("%d\n",step);
    return 0;
}

View Code

转载于:https://www.cnblogs.com/zero-begin/p/4313118.html