Fliptile(POJ 3279)

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• 原题如下：
  Fliptile

  Time Limit: 2000MS		Memory Limit: 65536K
  Total Submissions: 16494		Accepted: 6025

  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

• 题解：首先，同一个格子翻转两次的话就会恢复原状，所以多次反转是多余的，此外，翻转的格子的集合相同的话，其次序是无关紧要的，所以总共有2^MN种翻转的方法，由于解空间实在太大，我们必须另寻他径，参考解决POJ3276的方法，那道题中，让最左端的牛反转的方法只有1种，所以只要用直接判断的方法确定就可以了，但在这里，是行不通的，比如左上角的格子，除了翻转(1,1)之外，翻转(1,2)和(2,1)也都可以把(1,1)翻转，所以不能直接套用POJ3276的方法，但是如果假设第一行的翻转方法已经确定，那么翻转(1,1)的就只剩下(2,1)了，所以可以直接判断(2,1)是否需要翻转，类似的第二行都可以判断，如此反复下去就可以判断出所有格子的翻转方法了，判断是否有解，只要看最后一行是不是全为白色即可，如果并非全白，那么就说明不存在可行的操作方法。综上，我们只要先确定出第一行的翻转方式即可，而第一行的翻转方式共有2^N种，所以总的时间复杂度为O(MN2^N)
• 代码：

    1 #include <cstdio>
    2 #include <cctype>
    3 #include <algorithm>
    4 #include <cmath>
    5 #include <cstring>
    6 #define number s-‘0‘
    7 
    8 using namespace std;
    9 
   10 const int MAX_N=16;
   11 const int INF=0x3f3f3f3f;
   12 const int dx[5]={-1, 0, 0, 0, 1};
   13 const int dy[5]={0, 1, 0, -1, 0};
   14 int N,M;
   15 int flip[MAX_N][MAX_N], tile[MAX_N][MAX_N], opt[MAX_N][MAX_N];
   16 
   17 void read(int &x){
   18     char s;
   19     x=0;
   20     bool flag=0;
   21     while(!isdigit(s=getchar()))
   22         (s==‘-‘)&&(flag=true);
   23     for(x=number;isdigit(s=getchar());x=x*10+number);
   24     (flag)&&(x=-x);
   25 }
   26 
   27 void write(int x)
   28 {
   29     if(x<0)
   30     {
   31         putchar(‘-‘);
   32         x=-x;
   33     }
   34     if(x>9)
   35         write(x/10);
   36     putchar(x%10+‘0‘);
   37 }
   38 
   39 int calc();
   40 int get(int, int);
   41 
   42 int main()
   43 {
   44     read(M);read(N);
   45     for (int i=0; i<M; i++)
   46         for (int j=0; j<N; j++)
   47             read(j[i[tile]]);
   48     int res=-1;
   49     for (int i=0; i< 1<<N; i++)
   50     {
   51         memset(flip, 0, sizeof(flip));
   52         for (int j=0; j<N; j++)
   53         {
   54             flip[0][N-j-1]=i>>j&1;
   55         }
   56         int num=calc();
   57         if (num>=0 && (res<0 || num<res))
   58         {
   59             res=num;
   60             memcpy(opt, flip, sizeof(flip));
   61         }
   62     }
   63     if (res<0) puts("IMPOSSIBLE\n");
   64     else
   65         for (int i=0; i<M; i++)
   66             for (int j=0; j<N; j++)
   67                 printf("%d%c", j[i[opt]], j+1==N? ‘\n‘: ‘ ‘);
   68 }
   69 
   70 int get(int x, int y)
   71 {
   72     int c=tile[x][y];
   73     for (int i=0; i<5; i++)
   74     {
   75         int x2=x+dx[i], y2=y+dy[i];
   76         if (0<=x2 && x2<M && 0<=y2 && y2<N)
   77         {
   78             c+=y2[x2[flip]];
   79         }
   80     }
   81     return c % 2;
   82 }
   83 
   84 int calc()
   85 {
   86     for (int i=1; i<M; i++)
   87     {
   88         for (int j=0; j<N; j++)
   89         {
   90             if (get(i-1,j)!=0) flip[i][j]=1;
   91         }
   92     }
   93     for (int j=0; j<N; j++)
   94     {
   95         if (get(M-1,j)!=0) return -1;
   96     }
   97     int res=0;
   98     for (int i=0; i<M; i++)
   99         for (int j=0; j<N; j++)
  100             res+=j[i[flip]];
  101     return res;
  102 }

   

Fliptile(POJ 3279)

标签：cst   bsp   ips   ges   flag   lin   min   text   efi   

原文地址：https://www.cnblogs.com/Ymir-TaoMee/p/9509503.html