首页 > ACM题库 > HDU-杭电 > hdu 4328-cut the cake-动态规划-[解题报告]hoj
2015
05-23

hdu 4328-cut the cake-动态规划-[解题报告]hoj

http://acm.hdu.edu.cn/showproblem.php?pid=4328

Problem Description
Mark bought a huge cake, because his friend ray_sun’s birthday is coming. Mark is worried about how to divide the cake since it’s so huge and ray_sun is so strange. Ray_sun is a nut, you can never imagine how strange he was, is, and going to be. He does not
eat rice, moves like a cat, sleeps during work and plays games when the rest of the world are sleeping……It is not a surprise when he has some special requirements for the cake. A considering guy as Mark is, he will never let ray_sun down. However, he does
have trouble fulfilling ray_sun’s wish this time; could you please give him a hand by solving the following problem for him?
  The cake can be divided into n*m blocks. Each block is colored either in blue or red. Ray_sun will only eat a piece (consisting of several blocks) with special shape and color. First, the shape of the piece should be a rectangle. Second, the color of blocks
in the piece should be the same or red-and-blue crisscross. The so called ‘red-and-blue crisscross’ is demonstrated in the following picture. Could you please help Mark to find out the piece with maximum perimeter that satisfies ray_sun’s requirements?

 


Input
The first line contains a single integer T (T <= 20), the number of test cases.
  For each case, there are two given integers, n, m, (1 <= n, m <= 1000) denoting the dimension of the cake. Following the two integers, there is a n*m matrix where character B stands for blue, R red.
 


Output
For each test case, output the cased number in a format stated below, followed by the maximum perimeter you can find.
 


Sample Input
2 1 1 B 3 3 BBR RBB BBB
 


Sample Output
Case #1: 4 Case #2: 8
/**
hdu4328 最大子矩阵问题O(n*m)
题目大意:给定一个n*m的棋盘有红黑两色,让截取一个周长最大矩形,该矩形要么全是黑色,要么全是红色,要么黑色和红色交替
           见题目附图。
解题思路:如果遍历每一个矩阵是不可能的,时间O(n*n*n),因此我们采用扫描的方法(白书P51)。对于交替的矩形我们把(i+j)
           为奇数的格子翻转颜色,那样就可以转换成全是一样颜色了,然后边成把(i+j)为偶数的翻转,四种情况取最大即可。复杂度O(n*m)
*/
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;

const int maxn=1005;
int mat[maxn][maxn],mat1[maxn][maxn],up[maxn][maxn],lef[maxn][maxn],rig[maxn][maxn];
char a[maxn][maxn];
int n,m;
int get()
{
    int ans=0;
    for(int i=0; i<m; i++)
    {
        int lo=-1,ro=n;
        for(int j=0; j<n; j++)
        {
            if(mat[i][j]==1)
            {
                up[i][j]=lef[i][j]=0;
                lo=j;
            }
            else
            {
                up[i][j]=i==0?1:up[i-1][j]+1;
                lef[i][j]=i==0?lo+1:max(lef[i-1][j],lo+1);
            }
        }
        for(int j=n-1; j>=0; j--)
        {
            if(mat[i][j]==1)
            {
                rig[i][j]=n;
                ro=j;
            }
            else
            {
                rig[i][j]=i==0?ro-1:min(rig[i-1][j],ro-1);
                ans=max(ans,up[i][j]*2+2*(rig[i][j]-lef[i][j]+1));
            }
        }
    }
    return ans;
}
int main()
{
    int tt=0,T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&m,&n);
        for(int i=0; i<m; i++)
        {
            scanf("%s",a[i]);
        }
        for(int i=0; i<m; i++)
        {
            for(int j=0; j<n; j++)
            {
                if(a[i][j]=='R')
                    mat1[i][j]=mat[i][j]=1;
                else
                    mat1[i][j]=mat[i][j]=0;
            }
        }
        int ans=-1;
        ///1
        ans=max(ans,get());
        ///2
        for(int i=0; i<m; i++)
        {
            for(int j=0; j<n; j++)
            {
                if(mat1[i][j]==1)
                    mat[i][j]=0;
                else
                    mat[i][j]=1;
            }
        }
        ans=max(ans,get());
        ///3
        for(int i=0; i<m; i++)
        {
            for(int j=0; j<n; j++)
            {
                if((i+j)%2==1)
                {
                    if(mat1[i][j]==1)
                        mat[i][j]=0;
                    else
                        mat[i][j]=1;
                }
                else
                    mat[i][j]=mat1[i][j];
            }
        }
       /* for(int i=0; i<m; i++)
        {
            for(int j=0; j<n; j++)
            {
                printf("%d",mat[i][j]);
            }
            printf("\n");
        }
        printf("\n");*/
        ans=max(ans,get());
        ///4
        for(int i=0; i<m; i++)
        {
            for(int j=0; j<n; j++)
            {
                if((i+j)%2==0)
                {
                    if(mat1[i][j]==1)
                        mat[i][j]=0;
                    else
                        mat[i][j]=1;
                }
                else
                    mat[i][j]=mat1[i][j];
            }
        }
        /*for(int i=0; i<m; i++)
        {
            for(int j=0; j<n; j++)
            {
                printf("%d",mat[i][j]);
            }
            printf("\n");
        }
        printf("\n");*/
        ans=max(ans,get());
        printf("Case #%d: %d\n",++tt,ans);
    }
    return 0;
}
/**
100
3 3
BBR
RBR
BBB

2 2
BR
RB

2 2
BB
RR

2 2
BB
BB
*/

版权声明:本文为博主原创文章,未经博主允许不得转载。