HDU 2870

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

这题若先把w,x,y,z都转换成对应的字符(a,b,c)中的一种，则可转换成求由相同字符构成的最大矩阵，后面的做法就和和HDU 1505这题类似了。

h[i][j]表示以在某一行以第i列为底以第j种字符的最大高度。以此高度向左右分别扩散，找到高度大于等于当前高度的左边界lef[i]和右边界rig[i]，则以此高度构成的相同字符的最大矩阵的大小为：h[i][j]*(rig[i]-lef[i]);遍历每一行即可找到所求答案。

#include <cstdio>#include <cstring>#include <string>#include <queue>#include <stack>#include <vector>#include <map>#include <iostream>#include <algorithm>using namespace std;#define ll long long#define int64 __int64#define M 1005int n , m , h[M][3] , lef[M] , rig[M];char matrix[M][M];void Cal(int k , char ch)//对当前列的高度更新{if (ch == 'a'){h[k][0]++;h[k][1] = 0;h[k][2] = 0;}else if (ch == 'b'){h[k][0] = 0;h[k][1]++;h[k][2] = 0;}else if (ch == 'c'){h[k][0] = 0;h[k][1] = 0;h[k][2]++;}else if (ch == 'w'){h[k][0]++;h[k][1]++;h[k][2] = 0;}else if (ch == 'x'){h[k][0] = 0;h[k][1]++;h[k][2]++;}else if (ch == 'y'){h[k][0]++;h[k][1] = 0;h[k][2]++;}else{h[k][0]++;h[k][1]++;h[k][2]++;}}int Solve(){int i , j , l , ret = 0;memset(h , 0 , sizeof h);h[0][0] = h[0][1] = h[0][2] = -1;h[m+1][0] = h[m+1][1] = h[m+1][2] = -1;for (i = 1 ; i <= n ; i++){for (j = 1 ; j <= m ; j++){Cal(j,matrix[i][j]);}for (l =0 ; l < 3 ; l++){for (j = 1 ; j <= m ; j++){lef[j] = j;while (h[lef[j]-1][l] >= h[j][l])lef[j] = lef[lef[j]-1];}for (j = m ; j >= 1 ; j--){rig[j] = j;while (h[rig[j]+1][l] >= h[j][l])rig[j] = rig[rig[j]+1];}int MAX;for (j = 1 ; j <= m ; j++){MAX = h[j][l]*(rig[j]-lef[j]+1);ret = max(ret,MAX);}}}return ret;}int main(){while (~scanf("%d%d",&n,&m)){int i;for (i = 1 ; i <= n ; i++)scanf("%s",matrix[i]+1);printf("%d\n",Solve());}return 0;}