poj是什么意思，POJ 3279 Fliptile

傳送門：http://poj.org/problem?id=3279

Fliptile

Time Limit:?2000MS	?	Memory Limit:?65536K
Total Submissions:?8322	?	Accepted:?3102

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

//??現在還不會做，慢慢看吧，在網上看了一下他們的解題報告
1.如果處理了前面n-1行了，我們只要判斷最后一行是否是全為0
2.現在才明白怎們用暴力方法，每一個格子都有兩種操作 翻還是不翻，這樣就有2^(M+N)種可能性，然后一次枚舉每一種，找出翻轉最少的就是它的解。
3.還是不知道怎么弄，以后持續更新吧。
現在完全明白了，和POJ 3276 Face The Right Way  http://poj.org/problem?id=3276 類似，上面這個簡單點，
這類問題統稱為開關問題，它注重開頭和結尾，開頭決定了，后面的狀態也可由前面的狀態推出來，而判斷末尾情況，可以發現該翻轉方式是否可行。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

const int MAX_N=100;
const int dx[]={-1,1,0,0,0};
const int dy[]={0,0,-1,0,1};

int flip[MAX_N][MAX_N],tile[MAX_N][MAX_N],opt[MAX_N][MAX_N];
int N,M;

//這是判斷該瓷磚是否需要翻轉

int get(int x,int y){
    int c=tile[x][y];

    for(int i=0;i<5;i++){
        int tx=x+dx[i],ty=y+dy[i];
        if(tx>=0&&tx<M&&ty>=0&&ty<N)
            c+=flip[tx][ty];
    }
    return c%2;
}

int calc(){
    for(int i=1;i<M;i++)
    for(int j=0;j<N;j++)
    if(get(i-1,j)&1){    //如何判斷[i][j]瓷磚是否需要翻轉呢，這是判斷它的上一個[i-1][j]是否是1
        flip[i][j]=1;
    }

    //檢測最后一行是否全為0，如果不是則該翻轉不行。
    for(int j=0;j<N;j++){
        if(get(M-1,j)&1)
            return -1;
    }
    //這是計算翻轉了多少次
    int res=0;
    for(int i=0;i<M;i++)
        for(int j=0;j<N;j++)
        res+=flip[i][j];
    return res;
}

void solve(){

    int res=-1;
    //二進制枚舉第一行的所有選項
    for(int i=0;i<1<<N;i++){
        memset(flip,0,sizeof(flip));

        for(int j=0;j<N;j++)
            flip[0][N-j-1]=i>>j&1;

        int num=calc();

        if(num>=0&&(res<0||res>num)){
            memcpy(opt,flip,sizeof(flip));
            res=num;
        }
    }

    if(res<0) printf("IMPOSSIBLE\n");
    else {
        for(int i=0;i<M;i++)
            for(int j=0;j<N;j++)
             printf("%d%c",opt[i][j],j+1==N?'\n':' ');
    }
}

int main(){
    cin>>M>>N;
    for(int i=0;i<M;i++)
        for(int j=0;j<N;j++)
        scanf("%d",&tile[i][j]);
    solve();
}

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