Poj

Fliptile

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 10018		Accepted: 3717

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

USACO 2007 Open Silver

题目大意：略。

方法：枚举出第一行所有的状态，后面的状态都可以由第一行确定，因为如果上面那个位点有1，那么这个位点就必须翻转。一直到最后一行，看最后一行是否全为0。

输出操作次数最少的那个操作；如果操作次数一样，输出操作字典序最小的。

我用的是next_permutation枚举第一行操作的，ans数组是最后要输出的，一些细节写在代码里了。

代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
struct node
{
    int map[20][20];//把地图写成结构体，便于直接复制 
};
int n,m;
int dir[5][2]={
    0,0,1,0,0,1,-1,0,0,-1
};
int a[20][20];
int b[20];
node mm;//mm作为临时地图 
int ans[20][20];
int cnt=0,lastcnt=0;//cnt是翻转次数，lastcnt是上一次操作的翻转次数 
void flip(int x,int y)
{
    for(int i=0;i<5;i++)
    {
        int a=x+dir[i][0];
        int b=y+dir[i][1];
        if(mm.map[a][b])mm.map[a][b]=0;
        else mm.map[a][b]=1;
    }
}
int fun()
{
    for(int i=2;i<=m;i++)
    {
        for(int j=1;j<=n;j++)
        {
            if(mm.map[i-1][j])
            {
                a[i][j]=1;
                cnt++;
                flip(i,j);
            }
        }
    }
    for(int j=1;j<=n;j++)
    if(mm.map[m][j])return 0;//不能使地图最后一行全为0，返回0 
    return 1;//否则返回1 
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    node M;
    int flag=0;
    cin>>m>>n;
    for(int i=1;i<=m;i++)
    {
        for(int j=1;j<=n;j++)
        cin>>M.map[i][j];
    } 
    for(int i=0;i<=n;i++)//i表示第一行操作中有几个1 
    {
        memset(b,0,sizeof(b));
        for(int j=n-i+1;j<=n;j++)b[j]=1;//b表示第一行的操作 
        do
        {
            mm=M;
            cnt=i;//cnt表示翻转次数 
            memset(a,0,sizeof(a));
            for(int i=1;i<=n;i++)
            if(b[i])flip(1,i); 
            if(fun()&&(cnt<=lastcnt||lastcnt==0))//先比较cnt大小 
            {
                flag++;
                if(cnt<lastcnt||lastcnt==0)
                {
                    for(int i=1;i<=n;i++)ans[1][i]=b[i];
                    for(int i=2;i<=n;i++)
                    {
                        for(int j=1;j<=n;j++)
                        ans[i][j]=a[i][j];
                    }
                }
                else if(cnt==lastcnt)//如果cnt等于lastcnt，看字典序大小 
                {
                    bool f=false;
                    for(int i=1;i<n;i++)//比较字典序 
                    if(b[i]<ans[1][i]) 
                    {
                        f=true;
                        break;
                    }    
                    else if(b[i]>ans[1][i])break;
                    if(f)
                    {
                        for(int i=1;i<=n;i++)ans[1][i]=b[i];
                        for(int i=2;i<=n;i++)
                        {
                            for(int j=1;j<=n;j++)
                            ans[i][j]=a[i][j];
                        }
                    }
                }
                lastcnt=cnt;    //只有能使地图全为0的cnt才能更新lastcnt的值 ，也就是说写到花阔号外是错的 
            }
        }while(next_permutation(b+1,b+n+1));//用next_permutation枚举出所有的可能 
    }
    if(!flag)cout<<"IMPOSSIBLE"<<endl;
    else 
    {
        for(int i=1;i<=m;i++)
        {
            for(int j=1;j<n;j++)
            cout<<ans[i][j]<<" ";
            cout<<ans[i][n]<<endl;
        }
    }
    return 0;
}