斑的家

#include<iostream>
#include<cstring>
#define SIZE 10
/*
 * AUTHOR : benbendog
 * DATE   : 2010/12/18
 * PROBLEM: given a map of #,O (light off ,on) ,find the minimum switches
 *          to turn off all the O, once a switch at (i,j) is used,
 *          the values of 4 neighbors of (i,j) and (i,j) would flipped
 *          for example, ###                        #O#
 *                                #O#    flip (2,2)      O#O
 *                                 ###                        #O#
 * SOLUTION: The problem is which switch we have to turn off.
             when inspected row by row, row (i,j) would need to
 *           switch when row(i-1,j) is on, this is because
 *           we assume the inspected row is all off
 *           then the problem is the first row has nothing to
 *           depend on. Thus, we can generate all the possibilites
 *           of on/off for the first row (2^10)   
 */
using namespace std;
bool choose[SIZE];
bool t[SIZE][SIZE];
bool m[SIZE][SIZE];
int dx[]={1,-1,0,0,0};
int dy[]={0,0,1,-1,0};
int ans;
void switching(int x,int y);
void recur(int p);
bool complete();
int main()
{
    char ch[1000];
    while(cin>>ch && strcmp(ch,"end"))
    {
        int i,j;
        char c;
        for(i = 0 ;i < SIZE ; i++)
            for(j = 0 ;j < SIZE ;j++)
            {
                cin>>c;
                if(c=='#')m[i][j] = false;
                else      m[i][j] = true;
            }
        ans = 101;
        recur(0);
        
        cout<<ch<<' ';
        if(ans==101)cout<<-1<<endl;
        else         cout<<ans<<endl;
    }
    return 0;
}
void recur(int p)
{
    if(p==SIZE)
    {
        int temp = 0;
        memcpy(t,m,sizeof(m));
        for(int i = 0 ;i < SIZE ; i++)
        {
            if(choose[i]){switching(0,i);temp++;}
        }
        for(int i = 1 ;i < SIZE ; i++)
            for(int j = 0 ;j < SIZE ; j++)
                if(t[i-1][j])
                {
                    switching(i,j);temp++;
                }
        if(complete()){ans = min(ans,temp);}
    }
    else
    {
        choose[p] = true;
        recur(p+1);
        choose[p]= false;
        recur(p+1);
    }
}
bool complete()
{
    int i,j;
    for(i = 0 ;i < SIZE ; i++)
        for(j = 0 ;j < SIZE ;j++)
            if(t[i][j])return false;
    return true;
}
void switching(int x,int y)
{
    int tx,ty;
    for(int i = 0 ;i < 5 ; i++)
    {
        tx = x + dx[i];
        ty = y + dy[i];
        if(0<=tx && tx < SIZE && 0<=ty && ty < SIZE)t[tx][ty]=!t[tx][ty];
    }
}