hdu 5319 Painter 2015 Multi-University Training Contest 3

Painter

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 587 Accepted Submission(s): 275



Problem Description
Mr. Hdu is an painter, as we all know, painters need ideas to innovate , one day, he got stuck in rut and the ideas dry up, he took out a drawing board and began to draw casually. Imagine the board is a rectangle, consists of several
square grids. He drew diagonally, so there are two kinds of draws, one is like ‘\’ , the other is like ‘/’. In each draw he choose arbitrary number of grids to draw. He always drew the first kind in red color, and drew the other kind in blue color, when a
grid is drew by both red and blue, it becomes green. A grid will never be drew by the same color more than one time. Now give you the ultimate state of the board, can you calculate the minimum time of draws to reach this state.


Input
The first line is an integer T describe the number of test cases.

Each test case begins with an integer number n describe the number of rows of the drawing board.

Then n lines of string consist of ‘R’ ‘B’ ‘G’ and ‘.’ of the same length. ‘.’ means the grid has not been drawn.

1<=n<=50

The number of column of the rectangle is also less than 50.

Output

Output an integer as described in the problem description.



Output
Output an integer as described in the problem description.


Sample Input

2
4
RR.B
.RG.
.BRR
B..R
4
RRBB
RGGB
BGGR
BBRR

Sample Output

3
6

Source
2015 Multi-University Training Contest 3



Recommend
wange2014 | We have carefully selected several similar problems for you: 5326 5325 5324 5323 5322

#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<climits>
#include<queue>
#include<vector>
#include<map>
#include<sstream>
#include<set>
#include<stack>
#include<utility>
//#pragma comment(linker, "/STACK:102400000,102400000")
#define INF 0x3f3f3f3f
#define PI 3.1415926535897932384626
#define eps 1e-10
#define sqr(x) ((x)*(x))
#define FOR0(i,n)  for(int i=0 ;i<(n) ;i++)
#define FOR1(i,n)  for(int i=1 ;i<=(n) ;i++)
#define FORD(i,n)  for(int i=(n) ;i>=0 ;i--)
#define  lson   num<<1,l,mid
#define rson    num<<1|1,mid+1,r
#define MID   int mid=(l+r)>>1
#define zero(x)((x>0? x:-x)<1e-15)

using namespace std;
const int maxn=  50+5  ;
//const int maxm=    ;
//const int INF=    ;
//typedef long long ll;
//ifstream fin("input.txt");
//ofstream fout("output.txt");
//fin.close();
//fout.close();
//freopen("a.in","r",stdin);
//freopen("a.out","w",stdout);
char a[maxn][maxn];
int  red[maxn][maxn];
int  blue[maxn][maxn];
int n;int m;
bool in(int &y,int &x)
{
    return 1<=y&&y<=n&&1<=x&&x<=m;
}
/*
int solveblue(   )  我写的，虽然比较麻烦
{

    int ans=0;
    for(int stx=1;stx<=m+n-1;stx++)
    {
        int sty=1;
        bool has=0;
        for(int y=1,x=stx;y<=n;y++,x--)
        {
            if(!in(y,x) )  continue;
            if(blue[y][x]&&has==0)  {ans++; has=1;   }
            else if(!blue[y][x])
            {
                has=0;
            }

        }
    }

    return ans;
}

int solvered()
{
    int ans=0;
    for(int stx=2-n;stx<=m;stx++)
    {
        int sty=1;
        bool has=0;
        for(int y=sty,x=stx;y<=n;y++ ,x++ )
        {
            if(!in(y,x)  )  continue;
            if(red[y][x]&&has==0)  {has=1;ans++;}
            else if(!red[y][x])
            {
                has=0;
            }
        }
    }

    return ans;
}
*/

int solveblue()   //队友的找法，我写了一遍
{

    int ans=0;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            int y=i-1,x=j+1;
            if(blue[i][j]&&  (!in(y,x)||!blue[y][x])  )
                ans++;

        }
    }

    return ans;

}
int solvered()
{
    int ans=0;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            int y=i-1,x=j-1;
            if(red[i][j]&&  (!in(y,x)||!red[y][x])  )
                ans++;

        }
    }

    return ans;

}

int main()
{
      int i,j;
      int T;

      scanf("%d",&T);
      while(T--)
      {
          memset(blue,0,sizeof blue);
          memset(red,0,sizeof red);

          scanf("%d",&n);
          for(int i=1;i<=n;i++)
          {

                  scanf(" %s",a[i]+1);
                  m=strlen(a[i]+1);

                  for(int j=1;j<=m;j++)
              {
                if(a[i][j]=='G')  blue[i][j]=1,red[i][j]=1;
                else if(a[i][j]=='R')  red[i][j]=1;
                else if(a[i][j]=='B')  blue[i][j]=1;
              }

          }

          int ans=0;
//          cout<<"sd"<<endl;
         ans+= solveblue( );
          ans+=solvered();
//                    cout<<"sd"<<endl;

          printf("%d\n",ans);

      }

    return 0;
}
/*
2
4
RR.B
.RG.
.BRR
B..R
4
RRBB
RGGB
BGGR
BBRR

*/