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  • Painter's Problem poj1681 高斯消元法

    Painter's Problem
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 4420   Accepted: 2143

    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

      1 #include <iostream>
      2 #include <stdio.h>
      3 #include <algorithm>
      4 #include <math.h>
      5 #include <string.h>
      6 #include <set>
      7 using namespace std;
      8 const int MAXN=300;
      9 int a[MAXN][MAXN];
     10 int x[MAXN];
     11 bool free_x[MAXN];
     12 inline int gcd(int a,int b)
     13 {
     14     int t;
     15     while(b!=0)
     16     {
     17         t=b;
     18         b=a%b;
     19         a=t;
     20     }
     21     return a;
     22 }
     23 inline int lcm(int a,int b)
     24 {
     25     return a/gcd(a,b)*b;
     26 }
     27 int Gauss(int equ,int var)
     28 {
     29     int i,j,k;
     30     int max_r;
     31     int col;
     32     col=0;
     33     for(k = 0; k < equ && col < var; k++,col++)
     34     {
     35         max_r=k;
     36         for(i=k; i<equ; i++)
     37         {
     38             if(a[i][col])
     39             {
     40                 max_r=i;
     41                 break;
     42             }
     43         }
     44         if(max_r!=k)
     45         {
     46             for(j=k; j<var+1; j++) swap(a[k][j],a[max_r][j]);
     47         }
     48         if(a[k][col]==0)
     49         {
     50             k--;
     51             continue;
     52         }
     53         for(i=0; i<equ; i++)
     54         {
     55             if(i!=k&&a[i][col]!=0)
     56             {
     57                 for(j=0; j<var+1; j++)
     58                 {
     59                     a[i][j]^= a[k][j];
     60                 }
     61             }
     62         }
     63     }
     64     for (i = k; i < equ; i++)
     65     {
     66         if (a[i][col] != 0) return 0;
     67     }
     68     return 1;
     69 }
     70 int main()
     71 {
     72     int n,m,t,i,j;
     73     cin>>t;
     74     char x;
     75     while(t--)
     76     {
     77         memset(a,0,sizeof(a));
     78         cin>>n;
     79         m=n*n;
     80         for(i=0; i<m; i++)
     81         {
     82             if(i%n==0)getchar();
     83             x=getchar();
     84             if(x!='y')a[i][m]=1;
     85         }
     86         for(i=0; i<m; i++)
     87         {
     88             a[i][i]=1;
     89             if(i-n>=0)
     90                 a[i][i-n]=1;
     91             if(i+n<m)
     92                 a[i][i+n]=1;
     93             if(i%n)
     94                 a[i][i-1]=1;
     95             if((i+1)%n)
     96                 a[i][i+1]=1;
     97         }
     98         if(!Gauss(m,m))cout<<"inf"<<endl;
     99         else
    100         {
    101             int ans=0;
    102             for(i=0; i<m; i++)ans+=a[i][m]&1;
    103             cout<<ans<<endl;
    104         }
    105     }
    106 }
    View Code
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  • 原文地址:https://www.cnblogs.com/ERKE/p/3835418.html
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