zoukankan      html  css  js  c++  java
  • POJ3279(KB1-D 熄灯问题)

    Fliptile

    Description

    Farmer John knows that an intellectually satisfied cow is a happy cow who will give more milk. He has arranged a brainy activity for cows in which they manipulate an M × N grid (1 ≤ M ≤ 15; 1 ≤ N ≤ 15) of square tiles, each of which is colored black on one side and white on the other side.

    As one would guess, when a single white tile is flipped, it changes to black; when a single black tile is flipped, it changes to white. The cows are rewarded when they flip the tiles so that each tile has the white side face up. However, the cows have rather large hooves and when they try to flip a certain tile, they also flip all the adjacent tiles (tiles that share a full edge with the flipped tile). Since the flips are tiring, the cows want to minimize the number of flips they have to make.

    Help the cows determine the minimum number of flips required, and the locations to flip to achieve that minimum. If there are multiple ways to achieve the task with the minimum amount of flips, return the one with the least lexicographical ordering in the output when considered as a string. If the task is impossible, print one line with the word "IMPOSSIBLE".

    Input

    Line 1: Two space-separated integers: M and N 
    Lines 2..M+1: Line i+1 describes the colors (left to right) of row i of the grid with N space-separated integers which are 1 for black and 0 for white

    Output

    Lines 1..M: Each line contains N space-separated integers, each specifying how many times to flip that particular location.

    Sample Input

    4 4
    1 0 0 1
    0 1 1 0
    0 1 1 0
    1 0 0 1

    Sample Output

    0 0 0 0
    1 0 0 1
    1 0 0 1
    0 0 0 0

    Source

     
    枚举第一行。
     1 //2017-02-22
     2 #include <iostream>
     3 #include <cstdio>
     4 #include <cstring>
     5 
     6 using namespace std;
     7 
     8 bool grid[20][20], tmp[20][20];
     9 int n, m, ans[20], res[20];
    10 
    11 void flip(int x, int y)
    12 {
    13     tmp[x][y] = !tmp[x][y];
    14     if(x-1>=0)tmp[x-1][y] = !tmp[x-1][y];
    15     if(y-1>=0)tmp[x][y-1] = !tmp[x][y-1];
    16     tmp[x+1][y] = !tmp[x+1][y];
    17     tmp[x][y+1] = !tmp[x][y+1];
    18 }
    19 
    20 void solve()
    21 {
    22     int penn, minflip = 0x3f3f3f3f;
    23     bool fg = false;
    24     for(int i = 0; i < (1<<m); i++)
    25     {
    26         for(int x = 0; x < n; x++)
    27               for(int y = 0; y < m; y++)
    28                   tmp[x][y] = grid[x][y];
    29         for(int y = 0; y < m; y++)
    30             if(i&(1<<y))
    31                   flip(0, m-1-y);
    32         ans[0] = i;
    33         for(int x = 1; x < n; x++){
    34             penn = 0;
    35             for(int y = 0; y < m; y++){
    36                 if(tmp[x-1][y]){
    37                     flip(x, y);
    38                     penn += (1<<(m-1-y));
    39                 }
    40             }
    41             ans[x] = penn;
    42         }
    43         bool ok = true;
    44         for(int j = 0; j < m; j++)
    45               if(tmp[n-1][j])
    46                   ok = false;
    47         if(ok){
    48             fg = true;
    49             int cnt = 0;
    50             for(int j = 0; j < n; j++){
    51                 for(int pos = 0; pos < m; pos++)
    52                       if(ans[j]&(1<<(m-1-pos)))cnt++;
    53             }
    54             if(cnt < minflip){
    55                 minflip = cnt;
    56                 for(int k = 0; k < n; k++)
    57                       res[k] = ans[k];
    58             }
    59         }
    60     }
    61     if(!fg)cout<<"IMPOSSIBLE"<<endl;
    62     else{
    63         for(int j = 0; j < n; j++){
    64             for(int pos = 0; pos < m; pos++)
    65                   if(pos == m-1)cout<<(res[j]&(1<<(m-1-pos))?1:0)<<endl;
    66                 else cout<<(res[j]&(1<<(m-1-pos))?1:0)<<" ";
    67         }
    68     }
    69 }
    70 
    71 int main()
    72 {
    73     while(cin>>n>>m)
    74     {
    75         for(int i = 0; i < n; i++)
    76               for(int j = 0; j < m; j++)
    77                   cin>>grid[i][j];
    78         solve();
    79     }
    80 
    81     return 0;
    82 }
  • 相关阅读:
    2019年11月4日随堂测试 最多输入字母统计
    写增删改查中间遇到的问题
    2019年12月9日下午自习成果
    2019年12月16日 分级考试
    2019年11月18日 JAVA期中考试 增删改查
    sql语言积累
    【转载】Java项目中常用的异常处理情况总结
    泛型
    C#数字格式化输出
    委托,Lambda的几种用法
  • 原文地址:https://www.cnblogs.com/Penn000/p/6431255.html
Copyright © 2011-2022 走看看