poj 1681 Painter's Problem 高斯消元 枚举自由变元

Painter's Problem

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 5598 Accepted: 2705

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

Source

题意：类似于开灯问题

解法：高斯消元 , 需要枚举自由变元

#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>#include<vector>using namespace std;#define all(x) (x).begin(), (x).end()#define for0(a, n) for (int (a) = 0; (a) < (n); (a)++)#define for1(a, n) for (int (a) = 1; (a) <= (n); (a)++)#define sqr(x) ((x)*(x))#define ysk(x)  (1<<(x))typedef long long ll;typedef pair<int, int> pii;const int INF =0x3f3f3f3f;const int maxn=225    ;int n,equ,var;int x[maxn+10],freex[maxn+10];int a[maxn+10][maxn+10];int pos[maxn+10];void debug(void){    int i, j;    for (i = 0; i < equ; i++)    {        for (j = 0; j < var + 1; j++)        {            cout << a[i][j] << " ";        }        cout << endl;    }    cout << endl;}int guass(){    int row,col,free_num=0;//free_num 为自由变元数目//    debug();    for(row=0,col=0;row<equ&&col<var;row++,col++)//行指针row，列指针col，一个equ*row大小的矩阵    //其中equ为矩阵的行数    {   //第一步：进行初等行变换(交换行)，使得当前行当前列系数不为零         if(!a[row][col])         {             for(int i=row;i<equ;i++)             {                  if(a[i][col])                  {                      for(int k=col;k<=var;k++)                      {                          swap(a[i][k],a[row][k]);                      }                      break;                  }             }         }        //如果没有可以交换的行，那么当前列col对应的元为自由变元。指针继续又扫，而不会下降。         if(!a[row][col])  { row--;freex[free_num++]=col;   continue;}//第二步，为了使之变为阶梯形矩阵,将下方处于该列的系数全部消为0 (第i行减去第row行)         for(int i=row+1;i<equ;i++)  if(a[i][col])         {               for(int k=col;k<=var;k++)               {                   a[i][k]^=a[row][k];               }         }         pos[row]=col;//pos[row]记录了第row行第一个非零系数。    }    //第三步，判断无解的情况    //这个时候row行到下方的方程都是无用的，实际上矩阵的秩=row,就是第0到第row-1个方程。    for(int i=row;i<equ;i++)    {        if(a[i][var]) return -1;    }// 一般这里有个判断多解的情况。   // if(row!=col) return var-row;//整个过程是col不断右移，但是在某一段时间里，row可能会停住。    //因为这一行扫过去，a[i][col]持续为0    //自由变元的个数=变量数var-实际方程数row(矩阵的行秩)//该题目要求的是最小的步数，也就是∑xi最小，那么枚举自由变元，使得所有的xi之后最小。    int ans=INF;    for(int s=0;s<ysk(free_num);s++)    {        int cnt=0;        int ts=s;        for(int i=0;i<free_num;i++)        {            x[freex[i]]=ts&1;            if(x[freex[i]])  cnt++;            ts>>=1;        }        int j= var-1 ;        for(int i=row-1;i>=0;i--)        {            int p=pos[i];             x[p]=a[i][var];            for(int j= var-1;j>p;j--)            {                x[p]^=a[i][j]*x[j];            }            if(x[p])  cnt++;        }        ans=min(ans,cnt);    }   return ans;}void init(){    equ=sqr(n);    var=equ;    memset(a,0,sizeof a);    for0(i,equ)    {        a[i][i]=1;        if(i-n>=0)  a[i-n][i]=1;        if(i+n<equ) a[i+n][i]=1;        if(i%n !=0) a[i-1][i]=1;        if(i%n !=n-1) a[i+1][i]=1;    }}int main(){    int T;char ch;    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        init();        for0(i,equ)        {            scanf(" %c",&ch);            a[i][var]= (ch=='y')?0:1;//若原来是黄色，那么不需要变，为0。        }        int ans=guass();        if(ans==-1)  printf("inf\n");        else  printf("%d\n",ans);    }   return 0;}