poj 1681 Painter's Problem (高斯消元)

Painter's Problem

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 5079 Accepted: 2460

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

23yyyyyyyyy5wwwwwwwwwwwwwwwwwwwwwwwww

Sample Output

015

高斯消元，枚举自由变元

/*======================================================# Author: whai# Last modified: 2015-10-07 16:10# Filename: poj1681.cpp======================================================*/#include <iostream>#include <cstdio>#include <vector>#include <algorithm>#include <cstring>#include <string>#include <cmath>#include <set>#include <map>using namespace std;#define LL __int64#define PB push_back#define P pair<int, int>#define X first#define Y secondconst int N = 20 * 20;char str[N][N];LL inv(LL a, LL m) {LL p = 1, q = 0, b = m, c, d;while (b > 0) {c = a / b;d = a; a = b; b = d % b;d = p; p = q; q = d - c * q;}return p < 0 ? p + m : p;}const int G_MOD = 2;struct Gauss {int f[N][N]; //第一维为等式0~n-1, 第二维为变元0~m-1, m是解int x[N]; //解int f_x[N], f_tot; //自由变元的列号//n表示等式个数，m表示变元个数int gauss(int n, int m) {int i, c;for(i = 0, c = 0; i < n && c < m; ++c) {int k = i;while(k < n && !f[k][c]) ++k;if(f[k][c] == 0) {f_x[f_tot++] = c;continue;}for(int j = c; j <= m; ++j)swap(f[i][j], f[k][j]);for(int j = 0; j < n; ++j)if(i != j && f[j][c]) {int tmp = f[j][c] * inv(f[i][c], G_MOD) % G_MOD;for(int k = 0; k <= m; ++k) {f[j][k] = ((f[j][k] - f[i][k] * tmp) % G_MOD + G_MOD) % G_MOD;}}++i;}for(int j = i; j < n; ++j) if(f[j][m]) return -1;if(i < m) return m - i;for(int j = m - 1; j >= 0; --j) {x[j] = f[j][m] * inv(f[j][j], G_MOD) % G_MOD;}return 0;}void init() {f_tot = 0;memset(f, 0, sizeof(f));}}g;bool ok(int x, int y, int n) {if(x >= 0 && x < n && y >= 0 && y < n) return true;return false;}int dx[] = {-1, 1, 0, 0};int dy[] = {0, 0, -1, 1};void deal(int x, int y, int n) {int u = x * n + y;g.f[u][u] = 1;int st = 0;if(str[x][y] == 'w') st = 1;g.f[u][n * n] = st ^ 0;for(int i = 0; i < 4; ++i) {int tmpx = x + dx[i];int tmpy = y + dy[i];if(ok(tmpx, tmpy, n)) {int v = tmpx * n + tmpy;g.f[v][u] = 1;}}}void gao(int n) {g.init();for(int i = 0; i < n; ++i) {for(int j = 0; j < n; ++j) {deal(i, j, n);}}int flag = g.gauss(n * n, n * n);if(flag == -1) puts("inf");else if(flag == 0){int ans = 0;for(int i = 0; i < n * n; ++i) {ans += g.f[i][n * n];}cout<<ans<<endl;} else {int maxx = (1<<flag);int var = n * n;int ans = var;for(int x = 0; x < maxx; ++x) {int cnt = 0;for(int i = 0; i < flag; ++i) {if(x & (1 << i)) {g.x[g.f_x[i]] = 1;++cnt;} else {g.x[g.f_x[i]] = 0;}}for(int i = var - 1 - flag; i >= 0; --i) {int tmp = g.f[i][var];for(int j = i + 1; j < var; ++j) {if(g.f[i][j])tmp ^= g.x[j];}cnt += tmp;}ans = min(ans, cnt);}cout<<ans<<endl;}}int main() {int T;scanf("%d", &T);while(T--) {int n;scanf("%d", &n);for(int i = 0; i < n; ++i) {scanf("%s", str[i]);}gao(n);}return 0;}