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  • hdu 2870 Largest Submatrix

    Largest Submatrix

    http://acm.hdu.edu.cn/showproblem.php?pid=2870

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

    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
     
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    hdu 1505 加强版
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #define N 1001
    using namespace std;
    int n,m,k,minn,ans;
    char s[N+2][N+2],t[N+2][N+2];
    int l[N],r[N];
    int a[N],b[N];
    int q[N],tmp[N],head,tail;
    void change(char w,char x,char y,char z)
    {
        for(int i=1;i<=n;i++)
         for(int j=1;j<=m;j++)
          if(s[i][j]==x || s[i][j]==y || s[i][j]==z) t[i][j]=w;
          else t[i][j]=s[i][j]; 
    }
    void monotonous(int *c,int *d)
    {
        int h=1;
        while(h<=m && !c[h]) h++;
        if(h>m) return;
        q[0]=c[h]; tmp[0]=h;
        head=0; tail=1;
        for(int i=h+1;i<=m;i++)
        {
            if(!c[i]) while(head<tail) d[tmp[head++]]=i-1;
            else if(head==tail) q[tail]=c[i],tmp[tail++]=i;
            else
            {
                while(head<tail && c[i]<q[tail-1]) d[tmp[--tail]]=i-1;
                q[tail]=c[i];
                tmp[tail++]=i;
            }
        }
        while(head<tail) d[tmp[head++]]=tmp[tail-1];
    }
    void solve(char x)
    {
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                if(t[i][j]!=x) a[j]=b[m-j+1]=0;
                else
                {
                    if(t[i][j]==t[i-1][j]) a[j]++,b[m-j+1]++;
                     else a[j]=b[m-j+1]=1;
                }
            }
            monotonous(a,r);
            monotonous(b,l);
            for(int j=1;j<=m;j++) tmp[j]=l[j];
            for(int j=1;j<=m;j++) l[m-j+1]=m-tmp[j]+1;
            for(int j=1;j<=m;j++) ans=max(ans,(r[j]-l[j]+1)*a[j]);
        }
    }
    int main()
    {
        while(scanf("%d%d",&n,&m)!=EOF)
        {
            ans=0;
            for(int i=1;i<=n;i++) scanf("%s",s[i]+1);
            change('a','w','y','z'); solve('a');
            change('b','w','x','z'); solve('b');
            change('c','x','y','z'); solve('c');
            printf("%d
    ",ans);
        }
    }
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  • 原文地址:https://www.cnblogs.com/TheRoadToTheGold/p/7308821.html
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