【POJ 1681】 Painter's Problem （高斯消元）

【POJ 1681】 Painter's Problem （高斯消元）

Painter's Problem Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5119   Accepted: 2481 Description There is a square wall which is made of n*n small square bricks. Some bricks are white while some bricks are yellow. Bob is a painter and he wants to paint all the bricks yellow. But there is something wrong with Bob's brush. Once he uses this brush to paint brick (i, j), the bricks at (i, j), (i-1, j), (i+1, j), (i, j-1) and (i, j+1) all change their color. Your task is to find the minimum number of bricks Bob should paint in order to make all the bricks yellow.  Input The first line contains a single integer t (1 <= t <= 20) that indicates the number of test cases. Then follow the t cases. Each test case begins with a line contains an integer n (1 <= n <= 15), representing the size of wall. The next n lines represent the original wall. Each line contains n characters. The j-th character of the i-th line figures out the color of brick at position (i, j). We use a 'w' to express a white brick while a 'y' to express a yellow brick. Output For each case, output a line contains the minimum number of bricks Bob should paint. If Bob can't paint all the bricks yellow, print 'inf'. Sample Input 2 3 yyy yyy yyy 5 wwwww wwwww wwwww wwwww wwwww Sample Output 0 15 Source POJ Monthly--2004.06.27 张嘉龄

每个小格可以列出一个方程 对于一个格子只受到本身和上下左右共五个格子的影响 同时此格子最终只会是0/1（不翻/翻）以此初始化即可

高斯消元列出n*n个方程求解即可 要注意多解的情况枚举改变不定变元的值 找出翻动的最小步数

代码如下:

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>

using namespace std;
const int INF = 0x3f3f3f3f;

int a[233][233];
int x[233];
int free_x[233];
int free_num,equ,var;

void init(int n)
{
memset(a,0,sizeof(a));
memset(x,0,sizeof(x));
free_num = 0;
var = equ = n*n;

for(int i = 0; i < n; ++i)
{
for(int j = 0; j < n; ++j)
{
int t = i*n+j;
a[t][t] = 1;
if(i) a[(i-1)*n+j][t] = 1;
if(j) a[i*n+j-1][t] = 1;
if(i < n-1) a[(i+1)*n+j][t] = 1;
if(j < n-1) a[i*n+j+1][t] = 1;
}
}
}

int Gauss()
{
int k,col;
for(k = col = 0; k < equ && col < var; ++k, ++col)
{
int mx = k;
for(int i = k+1; i < equ; ++i)
{
if(abs(a[i][col]) > abs(a[mx][col])) mx = i;
}
if(!a[mx][col])
{
--k;
free_x[free_num++] = col;
continue;
}
if(mx != k)
{
for(int i = col; i <= var; ++i)
swap(a[k][i],a[mx][i]);
}
for(int i = k+1; i < equ; ++i)
{
if(a[i][col])
{
for(int j = col; j <= var; ++j)
a[i][j] ^= a[k][j];
}
}
}
for(int i = k; i < equ; ++i)
if(a[i][var]) return -1;
if(k < var) return var-k;
for(int i = var-1; ~i; --i)
{
x[i] = a[i][var];
for(int j = i+1; j < var; ++j)
x[i] ^= (a[i][j]&&x[j]);
}
return 0;
}

void solve()
{
int ans,cnt,id;
int t = Gauss();
if(t == -1) puts("inf");
else if(!t)
{
ans = 0;
for(int i = 0; i < var; ++i) ans += x[i];
printf("%d\n",ans);
}
else
{
ans = INF;
int tot = (1<<t);
for(int i = 0; i < tot; ++i)
{
cnt = 0;
for(int j = 0; j < t; ++j)
{
if(i&(1<<j))
{
x[free_x[j]] = 1;
cnt++;
}
else x[free_x[j]] = 0;
}
for(int j = var-t-1; ~j; --j)
{
for(id = j; id < var; ++id)
if(a[j][id]) break;
x[id] = a[j][var];
for(int k = id+1; k <var; ++k)
if(a[j][k]) x[id] ^= x[k];
cnt += x[id];
}

ans = min(ans,cnt);
}
printf("%d\n",ans);
}
}

int main()
{
char str[33];
int t,n;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
init(n);

for(int i = 0; i < n; ++i)
{
scanf("%s",str);
for(int j = 0; j < n; ++j)
{
if(str[j] == 'w') a[i*n+j][var] = 1;
else a[i*n+j][var] = 0;
}
}

solve();
}

return 0;
}