POJ3279 Fliptile

Fliptile
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 11319 Accepted: 4183
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
反轉開關+暴力枚舉

繼承反轉問題的基礎之上，這個操作需要對第一行的所有情況進行暴力枚舉。
我們盡可能的將影響往後放，而不要對前面進行影響。 那麽我們可以僅儅上方元素為1使 我們進行變動，這樣一來，就不會對之前的任何元素造成影響，最後也衹需要檢測最後一行是不是全都是0就可以判斷。
但是這樣一來 我們忽略掉了什麽呢？就是第一行元素的反轉 我們沒有進行過。所以衹需要對第一行進行情況的暴力枚舉pow(2,n)種情況，再按照上面的思路存一下最優情況就可以解決了。

AC代碼：

#include<iostream>#include<string>#include<algorithm>#include<cstring>#include<cmath>using namespace std;int a[20][20];int b[20][20];int v[20][20];int as[20][20];int k[20];int m, n;int MIN;bool flag = 0;void solve();int sum1(){    int t=0;    for (int i = 0; i < m; i++)    {        for (int j = 0; j < n; j++)        {            t += v[i][j];        }    }    return t;}int check(int i, int j){    if (i<0 || i>m || j<0 || j>n)        return 0;    return 1;}void back(int i,int j){    if (check(i, j))    {        if (a[i][j])        {            a[i][j] = 0;        }        else            a[i][j] = 1;    }}void change(int i, int j){    v[i][j]++;     back(i, j);    back(i - 1, j);    back(i + 1, j);    back(i, j + 1);    back(i, j - 1);}void getans(){    int ans = 0;    for (int i = 0; i < n; i++)    {        ans += a[m - 1][i];    }    if (ans == 0)    {        flag = 1;        if (sum1() < MIN)        {            MIN = sum1();            for (int i = 0; i < m; i++)            {                for (int j = 0; j < n; j++)                {                    as[i][j] = v[i][j];                }            }        }    }    //solve();}void gank(){    for (int i = 1; i < m; i++)    {        for (int j = 0; j < n; j++)        {            if (a[i - 1][j] == 1)            {                change(i, j);            }        }    }}void solve(){    if (flag)    {        for (int i = 0; i < m; i++)        {            for (int j = 0; j < n; j++)            {                if (j)                    cout << " ";                cout << as[i][j];            }            cout << endl;        }    }    else    {        cout << "IMPOSSIBLE" << endl;    }}int main(){    while (cin >> m >> n)    {        MIN = 0x3f3f3f3f;        flag = 0;        int t =1<<n;        memset(a, 0, sizeof(a));        memset(v, 0, sizeof(v));        memset(as, 0, sizeof(as));        memset(b, 0, sizeof(b));        for (int i = 0; i < m; i++)        {            for (int j = 0; j < n; j++)            {                cin >> a[i][j];                b[i][j] = a[i][j];            }        }        for (int loop = 0; loop < t; loop++)         {            for (int i = 0; i < m; i++) //更新数组a            {                for (int j = 0; j < n; j++)                 {                    a[i][j]=b[i][j];                }            }            memset(v, 0, sizeof(v));            int t = loop;            for (int i = 0; i < n; i++)            {                k[i] = t % 2;                t /= 2;            }            for (int i = 0; i < n; i++)            {                if (k[i])                    change(0, i);            }            gank();            getans();        }        solve();    }    return 0;}