zoukankan      html  css  js  c++  java
  • HDU 2870 Largest Submatrix

    Largest Submatrix

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 839    Accepted Submission(s): 407

    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 4
    abcw
    wxyz
     
    Sample Output
    3
     
    Source
     
    Recommend
    gaojie
     
     
     
     
    #include<stdio.h>
    #include<string.h>
    
    const int maxn=1010;
    
    char str[maxn][maxn];
    int dp[maxn][maxn];
    int m,n,ans,l[maxn],r[maxn];
    
    void Solve(){
        int i,j;
        for(i=1;i<=m;i++){
            dp[i][0]=dp[i][n+1]=-1;
            for(j=1;j<=n;j++){
                l[j]=j;
                while(dp[i][j]<=dp[i][l[j]-1])
                    l[j]=l[l[j]-1];
            }
            for(j=n;j>=1;j--){
                r[j]=j;
                while(dp[i][j]<=dp[i][r[j]+1])
                    r[j]=r[r[j]+1];
            }
            for(j=1;j<=n;j++)
                if(ans<(r[j]-l[j]+1)*dp[i][j])
                    ans=(r[j]-l[j]+1)*dp[i][j];
        }
    }
    
    int main(){
    
        //freopen("input.txt","r",stdin);
    
        memset(dp,0,sizeof(dp));
        while(scanf("%d%d",&m,&n)!=EOF){
            //getchar();
            int i,j;
            ans=0;
            for(i=1;i<=m;i++)
                scanf("%s",str[i]+1);
            //for(i=1;i<=m;i++)
                //printf("%s\n",str[i]+1);
            //printf("m=%d n=%d\n",m,n);
            for(i=1;i<=m;i++)
                for(j=1;j<=n;j++){
                    if(str[i][j]=='a' || str[i][j]=='w' || str[i][j]=='y' || str[i][j]=='z')
                        dp[i][j]=dp[i-1][j]+1;
                    else
                        dp[i][j]=0;
                }
            //printf("m=%d n=%d\n",m,n);
            Solve();
            //printf("m=%d n=%d\n",m,n);
            for(i=1;i<=m;i++)
                for(j=1;j<=n;j++){
                    if(str[i][j]=='b' || str[i][j]=='w' || str[i][j]=='x' || str[i][j]=='z')
                        dp[i][j]=dp[i-1][j]+1;
                    else
                        dp[i][j]=0;
                }
            Solve();
            for(i=1;i<=m;i++)
                for(j=1;j<=n;j++){
                    if(str[i][j]=='c' || str[i][j]=='x' || str[i][j]=='y' || str[i][j]=='z')
                        dp[i][j]=dp[i-1][j]+1;
                    else
                        dp[i][j]=0;
                }
            Solve();
            printf("%d\n",ans);
        }
        return 0;
    }
  • 相关阅读:
    Windows 代码实现关机(直接黑屏)
    Windows SEH学习 x86
    Smali 语法文档
    SIOCADDRT: No such process
    Windbg 常用命令整理
    ida GDB 远程调试
    IDA 使用技巧
    Windows X64 Patch Guard
    C 和C++ 名称修饰规则
    【转载】 硬盘主引导记录(MBR)及其结构详解
  • 原文地址:https://www.cnblogs.com/jackge/p/2980706.html
Copyright © 2011-2022 走看看