hdu 2870 Largest Submatrixhdu

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

题意：一个矩形由字符a、b、c、w、x、y、z组成，其中w可以代替a、b，x可以代替b、c，y可以代替a、c，z可以代替a、b、c。求由相同组成的最大内矩形面积。

 思路：方法与hdu 1505一样，不过需要开三个数组a[]、b[]、c[]保存相同字符的区域。

  1 #include <cstdio>
  2 #include <cstring>
  3 #define N 1005
  4 
  5 int a[N][N], b[N][N], c[N][N], ha[N][N], hb[N][N], hc[N][N], l[N], r[N];
  6 char ss[N];
  7 int max2(int x, int y)
  8 {
  9     return x > y ? x : y;
 10 }
 11 
 12 int main()
 13 {
 14     int n, m;
 15     while(scanf("%d%d",&n, &m)!=EOF)
 16     {
 17         memset(a, 0, sizeof(a));
 18         memset(b, 0, sizeof(b));
 19         memset(c, 0, sizeof(c));
 20         memset(ha, 0, sizeof(ha));
 21         memset(hb, 0, sizeof(hb));
 22         memset(hc, 0, sizeof(hc));
 23         for(int i=1; i<=n; i++)
 24         {
 25             scanf("%s",ss+1);
 26             for(int j=1; j<=m; j++)
 27             {
 28                 if(ss[j]=='a')
 29                     a[i][j] = 1;
 30                 else if(ss[j]=='b')
 31                     b[i][j] = 1;
 32                 else if(ss[j]=='c')
 33                     c[i][j] = 1;
 34                 else if(ss[j]=='w')
 35                 {
 36                     a[i][j] = 1;
 37                     b[i][j] = 1;
 38                 }
 39                 else if(ss[j]=='x')
 40                 {
 41                     b[i][j] = 1;
 42                     c[i][j] = 1;
 43                 }
 44                 else if(ss[j]=='y')
 45                 {
 46                     a[i][j] = 1;
 47                     c[i][j] = 1;
 48                 }
 49                 else if(ss[j]=='z')
 50                 {
 51                     a[i][j] = 1;
 52                     b[i][j] = 1;
 53                     c[i][j] = 1;
 54                 }
 55             }
 56         }
 57 
 58 
 59         //a
 60         int ans = 0;
 61         for(int i=1; i<=n; i++)
 62         {
 63             for(int j=1; j<=m; j++)
 64             {
 65                 l[j] = r[j] = j;
 66                 if(a[i][j]==1) ha[i][j] = ha[i-1][j] + 1;
 67             }
 68             for(int j=1; j<=m; j++)
 69             {
 70                 while(l[j]-1>=1 && ha[i][l[j]-1]>=ha[i][j])
 71                     l[j] = l[l[j]-1];
 72             }
 73             for(int j=m-1; j>=1; j--)
 74             {
 75                 while(r[j]+1<=m && ha[i][r[j]+1]>=ha[i][j])
 76                     r[j] = r[r[j]+1];
 77             }
 78             for(int j=1; j<=m; j++)
 79                 ans = max2(ans, (r[j]-l[j]+1)*ha[i][j]);
 80             //b
 81             for(int j=1; j<=m; j++)
 82             {
 83                 l[j] = r[j] = j;
 84                 if(b[i][j]==1) hb[i][j] = hb[i-1][j] + 1;
 85             }
 86             for(int j=1; j<=m; j++)
 87             {
 88                 while(l[j]-1>=1 && hb[i][l[j]-1]>=hb[i][j])
 89                     l[j] = l[l[j]-1];
 90             }
 91             for(int j=m-1; j>=1; j--)
 92             {
 93                 while(r[j]+1<=m && hb[i][r[j]+1]>=hb[i][j])
 94                     r[j] = r[r[j]+1];
 95             }
 96             for(int j=1; j<=m; j++)
 97                 ans = max2(ans, (r[j]-l[j]+1)*hb[i][j]);
 98             //c
 99             for(int j=1; j<=m; j++)
100             {
101                 l[j] = r[j] = j;
102                 if(c[i][j]==1) hc[i][j] = hc[i-1][j] + 1;
103             }
104             for(int j=1; j<=m; j++)
105             {
106                 while(l[j]-1>=1 && hc[i][l[j]-1]>=hc[i][j])
107                     l[j] = l[l[j]-1];
108             }
109             for(int j=m-1; j>=1; j--)
110             {
111                 while(r[j]+1<=m && hc[i][r[j]+1]>=hc[i][j])
112                     r[j] = r[r[j]+1];
113             }
114             for(int j=1; j<=m; j++)
115                 ans = max2(ans, (r[j]-l[j]+1)*hc[i][j]);
116         }
117         printf("%d\n",ans);
118     }
119     return 0;
120 }