HDOJ题目2870 Largest Submatrix（动态规划）

Largest Submatrix

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1562    Accepted Submission(s): 745

Problem Description

Now here is a matrix with letter 'a','b','c','w','x','y','z' and you can change 'w' to 'a' or 'b', change 'x' to 'b' or 'c', change 'y' to 'a' or 'c', and change 'z' to 'a', 'b' or 'c'. After you changed it, what's the largest submatrix with the same letters you can make?

 

Input

The input contains multiple test cases. Each test case begins with m and n (1 ≤ m, n ≤ 1000) on line. Then come the elements of a matrix in row-major order on m lines each with n letters. The input ends once EOF is met.

 

Output

For each test case, output one line containing the number of elements of the largest submatrix of all same letters.

 

Sample Input

2 4abcwwxyz

 

Sample Output

3

 

Source

2009 Multi-University Training Contest 7 - Host by FZU

 

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ac代码

#include<stdio.h>#include<string.h>int dp[1010][1010],l[1010],r[1010];char s[1010][1010];int n,m,max;void fun(){int i,j;for(i=1;i<=n;i++){for(j=1;j<=m;j++){l[j]=r[j]=j;}dp[i][0]=dp[i][m+1]=-1;for(j=2;j<=m;j++){while(dp[i][j]<=dp[i][l[j]-1])l[j]=l[l[j]-1];}for(j=m-1;j>=1;j--){while(dp[i][j]<=dp[i][r[j]+1])r[j]=r[r[j]+1];}for(j=1;j<=m;j++){int res=(r[j]-l[j]+1)*dp[i][j];if(res>max)max=res;}}}int main(){//int n,m;while(scanf("%d%d",&n,&m)!=EOF){int i,j;max=0;for(i=1;i<=n;i++){scanf("%s",s[i]);}for(i=1;i<=n;i++){for(j=1;j<=m;j++){if(s[i][j]=='a'||s[i][j]=='w'||s[i][j]=='y'||s[i][j]=='z'){dp[i][j]=dp[i-1][j]+1;}elsedp[i][j]=0;}}fun();for(i=1;i<=n;i++){for(j=1;j<=m;j++){if(s[i][j]=='b'||s[i][j]=='w'||s[i][j]=='x'||s[i][j]=='z'){dp[i][j]=dp[i-1][j]+1;}elsedp[i][j]=0;}}fun();for(i=1;i<=n;i++){for(j=1;j<=m;j++){if(s[i][j]=='c'||s[i][j]=='y'||s[i][j]=='x'||s[i][j]=='z'){dp[i][j]=dp[i-1][j]+1;}elsedp[i][j]=0;}}fun();printf("%d\n",max);}}