HDU 4328 Cut the cake

Cut the cake

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1250    Accepted Submission(s): 489

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

Author

BJTU

Source

2012 Multi-University Training Contest 3

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 #include<iostream>
 #include<cstring>
 #include<cstdio>
 using namespace std;
 #include<algorithm>
 #define N 1010
 int n,m,h[N][N],l[N][N],r[N][N];
 char g[N][N];
 int ans=,tmpl[N],tmpr[N];
 void solve(char key){
     memset(h,,sizeof(h));
     memset(l,,sizeof(l));
     memset(r,,sizeof(r));
     int mx=;
     for(int i=;i<=m;i++){
         l[][i]=;r[][i]=m;
     }
     for(int i=;i<=n;i++){
         int tmp=;
         for(int j=;j<=m;j++){
             if(g[i][j]!=key)tmp=j+;
             else tmpl[j]=tmp;
         }
         tmp=m;
         for(int j=m;j>=;j--){
             if(g[i][j]!=key) tmp=j-;
             else tmpr[j]=tmp;
         }
         //记录好了一个点能到达的 最左边 和最右边,
         //接下来就是dp了
         for(int j=;j<=m;j++){
             if(g[i][j]!=key){
                 l[i][j]=;h[i][j]=;r[i][j]=m;
                 continue;
             }
             h[i][j]=h[i-][j]+;
             l[i][j]=max(l[i-][j],tmpl[j]);
             r[i][j]=min(r[i-][j],tmpr[j]);
             mx=max(*(r[i][j]-l[i][j]++h[i][j]),mx);
         }
     }
     ans=max(ans,mx);
 }
 int main()
 {
     int T,tt=;scanf("%d",&T);
     for(int tz=;tz<=T;tz++){
         ans=;
         scanf("%d%d",&n,&m);
         for(int i=;i<=n;i++)
           scanf("%s",g[i]+);
         solve('B');solve('R');
         for(int i=;i<=n;i++)
           for(int j=;j<=m;j++)
             if((i+j)%==){
               if(g[i][j]=='B')g[i][j]='R';
               else g[i][j]='B';
             }
         solve('B');solve('R');
         printf("Case #%d: ",tt++);
         printf("%d\n",ans);
     }
     return ;
 }

思路：悬线法~~~~DP求满足题意的最大子矩阵的边长