POJ 3279 （状态压缩暴力枚举）

Fliptile

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 6357		Accepted: 2424

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

题意：给你一个M*N的数字矩阵，数字是0或1，让你反转数次使得整个矩阵所有数字都为0。反转规则是：当你反转一个数字，其相邻的数字也会被反转。问最少需要反转几次，输出每个点的反转次数。

题解：好吧，看到这个题完全懵逼了，这里根据题意可以先枚举第1行的所有反转状态，使用状态压缩的思想枚举第一行的所有可能性，然后就是枚举了，如果数字矩阵里

G[I-1][J]==1，那么就需要反转G[I][J]这个数字块了

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<string>
#include<bitset>
#include<utility>
#include<functional>
#include<iomanip>
#include<sstream>
#include<ctime>
using namespace std;

#define N int(1e1+6)
#define inf int(0x3f3f3f3f)
#define mod int(1e9+7)
typedef long long LL;

int g

,mp

,f

,cnt,n,m;

void flip(int x,int y)
{
cnt++;
mp[x][y]=1;
f[x][y]^=1;
if(x<n-1)f[x+1][y]^=1;
if(x>0)f[x-1][y]^=1;
if(y<m-1)f[x][y+1]^=1;
if(y>0)f[x][y-1]^=1;
}

bool ok(int x)
{
cnt=0;
memcpy(f,g,sizeof(f));
for(int i=0;i<m;i++)
{
if(x&(1<<(m-i-1)))
{
flip(0,i);
}
}
for(int i=1;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(f[i-1][j])
flip(i,j);
}
}
for(int i=0;i<m;i++)
{
if(f[n-1][i])return false;
}
return true;
}

int main()
{
#ifdef CDZSC
freopen("i.txt", "r", stdin);
//freopen("o.txt","w",stdout);
int _time_jc = clock();
#endif
while(~scanf("%d%d",&n,&m))
{
int ans=inf,way=-1;
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
scanf("%d",&g[i][j]);
}
}
for(int i=0;i<(1<<m);i++)
{
if(ok(i)&&ans>cnt)
{
ans=cnt;
way=i;
}
}
memset(mp,0,sizeof(mp));

if(ans!=inf)
{
ok(way);
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(j)printf(" ");
printf("%d",mp[i][j]);
}
puts("");
}
}
else
{
puts("IMPOSSIBLE");
}
}
return 0;
}