【最大子矩阵】hdu 2870

该题首先转化为只含有a，b，c的矩阵，然后对每一行dp求该列连续相同字母的最大高度，然后就可以每行每行地gao，又转化为求最大矩阵面积，就像hdu 1506一样，迭代地求左右>=当前高度的下标，hdu 1506代码看这里

#include <list>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <deque>
#include <stack>
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <limits.h>
#include <time.h>
#include <string.h>
using namespace std;

#define LL long long
#define PI acos(-1.0)
#define Max INT_MAX
#define Min INT_MIN
#define eps 1e-8
#define FRE freopen("a.txt","r",stdin)
#define N 1010
int max(int a,int b){return a>b?a:b;}
char str

;
int dp

;
int l
,r
;
int n,m;
int ans;
void gao(){
int i,j,k;
for(i=1;i<=n;i++){
for(j=1;j<=m;j++){
l[j]=j;
r[j]=j;
}
dp[i][0]=dp[i][m+1]=-1;
for(j=1;j<=m;j++)
while(dp[i][l[j]-1] >= dp[i][j])
l[j]=l[l[j]-1];

for(j=m;j>=1;j--)
while(dp[i][r[j]+1] >= dp[i][j])
r[j]=r[r[j]+1];

for(j=1;j<=m;j++)
ans=max(ans,(r[j]-l[j]+1)*dp[i][j]);
}
}
int main(){
while(scanf("%d%d",&n,&m)!=EOF){
int i,j;
for(i=1;i<=n;i++)scanf("%s",str[i]+1);
for(i=0;i<=m;i++)dp[0][i]=0;
ans=0;
for(i=1;i<=n;i++){
for(j=1;j<=m;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;

}
}
gao();

for(i=1;i<=n;i++){
for(j=1;j<=m;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;
}
}
gao();

for(i=1;i<=n;i++){
for(j=1;j<=m;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;
}
}
gao();

printf("%d\n",ans);
}
return 0;
}