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  • poj 1753 Flip Game

    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: 
    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

    题目大意:就是给你一个4*4的棋盘,通过翻转棋子来让整个棋盘变为全黑或全白,问几步可完成操作。
    注意:翻转其某个棋子时,其上下左右都要做相应的翻转,即黑变白,白变黑。
     1 #include <iostream>
     2 #include <stdio.h>
     3 using namespace std;
     4 bool chess[6][6]= {{false}};
     5 int r[6]= {0,0,0,-1,1};
     6 int c[6]= {0,1,-1,0,0};
     7 int step;
     8 bool flag;
     9 bool reach_all()   //判断是否为一种颜色
    10 {
    11     for(int i=1;i<5;i++)
    12     {
    13         for(int j=1;j<5;j++)
    14         {
    15             if(chess[i][j]!=chess[1][1])
    16                 return false;
    17         }
    18     }
    19     return true;
    20 }
    21 void turn(int row,int col) //翻棋
    22 {
    23     for(int i=0;i<5;i++)
    24         chess[row+r[i]][col+c[i]]=!chess[row+r[i]][col+c[i]];
    25     return ;
    26 }
    27 void dfs(int row,int col,int deep) //  深搜迭代回溯很重要
    28 {
    29     if(deep==step)
    30     {
    31         flag=reach_all();
    32         return ;
    33     }
    34     if(flag||row==5)
    35         return ;
    36     turn(row,col); //翻棋
    37     if(col<4)
    38         dfs(row,col+1,deep+1);
    39     else
    40         dfs(row+1,1,deep+1);
    41     turn(row,col);  //不符合则翻回来
    42     if(col<4)
    43         dfs(row,col+1,deep);
    44     else
    45         dfs(row+1,1,deep);
    46     return ;
    47 }
    48 int main()
    49 {
    50     char ch;
    51     for(int i=1; i<5; i++)
    52     {
    53         for(int j=1; j<5; j++)
    54         {
    55             scanf("%c",&ch);
    56             if(ch=='b')
    57                 chess[i][j]=true;
    58         }
    59         getchar();   //之前忘了写,就一直输出Impossible
    60     }
    61     for(step=0; step<=16; step++) //枚举每一种情况
    62     {
    63         dfs(1,1,0);
    64         if(flag)
    65             break;
    66     }
    67     if(flag)
    68         printf("%d
    ",step);
    69     else
    70         printf("Impossible
    ");
    71     return 0;
    72 }
    View Code
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  • 原文地址:https://www.cnblogs.com/cxbky/p/4823518.html
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