zoj2977Strange Billboard (国家压缩+罗列)

Strange Billboard

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Time Limit: 2 Seconds
Memory Limit: 65536 KB

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The marketing and public-relations department of the Czech Technical University has designed a new reconfigurable mechanical Flip-Flop Bill-Board (FFBB). The billboard is a regular two-dimensional grid of
R * C square tiles made of plastic. Each plastic tile is white on one side and black on the other. The idea of the billboard is that you can create various pictures by flipping individual tiles over. Such billboards will hang above all entrances
to the university and will be used to display simple pictures and advertise upcoming academic events.

To change pictures, each billboard is equipped with a "reconfiguration device". The device is just an ordinary long wooden stick that is used to tap the tiles. If you tap a tile, it flips over to the other side, i.e., it changes from white to black or vice
versa. Do you agree this idea is very clever?

Unfortunately, the billboard makers did not realize one thing. The tiles are very close to each other and their sides touch. Whenever a tile is tapped, it takes all neighboring tiles with it and all of them flip over together. Therefore, if you want to change
the color of a tile, all neighboring tiles change their color too. Neighboring tiles are those that touch each other with the whole side. All inner tiles have 4 neighbors, which means 5 tiles are flipped over when tapped. Border tiles have less neighbors,
of course.

name=0000%2F2977%2Fbillboard_0.jpg" alt="">

For example, if you have the billboard configuration shown in the left picture above and tap the tile marked with the cross, you will get the picture on the right. As you can see, the billboard reconfiguration is not so easy under these conditions. Your
task is to find the fastest way to "clear" the billboard, i.e., to flip all tiles to their white side.

Input

The input consists of several billboard descriptions. Each description begins with a line containing two integer numbers
R and C (1 <= R,C <= 16) specifying the billboard size. Then there are
R lines, each containing C characters. The characters can be either an uppercase letter "X" (black) or a dot "." (white). There is one empty line after each map. The input is terminated by two zeros in place of the board size.

Output

For each billboard, print one line containing the sentence "You have to tap T tiles.", where T is the minimal possible number of taps needed to make all squares white. If the situation cannot be solved, output the string "Damaged billboard." instead.

Sample Input

5 5
XX.XX
X.X.X
.XXX.
X.X.X
XX.XX

5 5
.XX.X
.....
..XXX
..X.X
..X..

1 5
...XX

5 5
...X.
...XX
.XX..
..X..
.....

8 9
..XXXXX..
.X.....X.
X..X.X..X
X.......X
X.X...X.X
X..XXX..X
.X.....X.
..XXXXX..

0 0

Sample Output

You have to tap 5 tiles.
Damaged billboard.
You have to tap 1 tiles.
You have to tap 2 tiles.
You have to tap 25 tiles.

题意：用最少的翻转把全部的X变成点。每翻第（x,y）点同一时候也会带动周围四个点。
解法：枚举第一行的翻转状态，依据第i-1行翻转第i行。

#include<stdio.h>
#include<iostream>
using namespace std;
#define mulit(j) (1<<j)
#define inf 999999999
int row[18],trow[18],n,m;
int dfs(int i,int prerow,int step)
{
    if(i==n)
    {
        if(prerow)return inf; else return step;
    }
    int now=trow[i],j;
    for(j=0;j<m;j++)
    if(prerow&(1<<j))
    {
        step++;  now^=mulit(j);
        if(j>0)now^=mulit(j-1);
        if(j<m-1)now^=mulit(j+1);
        if(i+1<n)trow[i+1]^=mulit(j);
    }
   return dfs(i+1,now,step);
}
void answer()
{
    int MIN=inf,step;
    for(int i=0;i<(1<<m);i++)
    {
        step=0;
        for(int j=0;j<n;j++)
        trow[j]=row[j];
        for(int j=0;(1<<j)<=i;j++)
        if(i&(1<<j))
        {
            step++; trow[0]^=(1<<j);
            if(j>0)trow[0]^=(1<<(j-1));
            if(j<m-1)trow[0]^=(1<<(j+1));
            if(n>1) trow[1]^=(1<<j);
        }
        step=dfs(1,trow[0],step);
        if(step<MIN)MIN=step;
    }
    if(MIN==inf)printf("Damaged billboard.\n");
    else printf("You have to tap %d tiles.\n",MIN);
}
int main()
{
    char ss[18];
    while(scanf("%d%d",&n,&m)>0&&n+m!=0)
    {
        for(int i=0;i<n;i++)
        {
            scanf("%s",ss);
            row[i]=0;
            for(int j=0;j<m;j++)
            if(ss[j]=='X')
                row[i]|=(1<<j);
        }
       answer();
    }
}

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