Different Rectangle

Different Rectangle

Description

Input

Output

Sample Input 1 

1
2 3
HPU
UHP

Sample Output 1

6

#pragma GCC optimize ("O3")
#pragma GCC optimize ("O2")
#include <bits/stdc++.h>
#include <ext/rope>
using namespace std;
using namespace __gnu_cxx;
#define met(s) memset(s, 0, sizeof(s))
#define rep(i, a, b) for(int i = a; i <= b; ++i)
template <class T> inline void scan_d(T &ret) {
char c; ret = 0;
while ((c = getchar()) < '0' || c > '9');
while (c >= '0' && c <= '9') { 
ret = ret * 10 + (c - '0'), c = getchar();}}
typedef long long LL;
typedef unsigned long long ull;
typedef pair<int, int> pii;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int MAXN = 1e2 + 10;
char str[MAXN][MAXN];
int dp1[MAXN][MAXN], dp2[MAXN][MAXN], vis[30];

int main() {
	// freopen("in.txt", "r", stdin);
 //    freopen("out.txt", "w", stdout);
	int T;
	scanf("%d", &T);
	while(T--) {
		int n, m;
		scanf("%d %d", &n, &m);
		for(int i = 0; i < n; ++i) scanf("%s", str[i]);
		for(int i = 0; i < n; ++i) {
			for(int j = 0; j < m; ++j) {
				dp1[i][j] = dp2[i][j] = 0;
				memset(vis, 0, sizeof(vis));
				for(int p = 0; ;++p) {
					if(vis[str[i][j + p] - 'A'] || j + p >= m) break;
					vis[str[i][j + p] - 'A'] = 1;
					dp1[i][j]++;//第i行（j向右推进）从dp1[i][j] 向右没有重复的个数 
				}
				memset(vis, 0, sizeof(vis));
				for(int p = 0; ;++p) {
					if(vis[str[i + p][j] - 'A'] || i + p >= n) break;
					vis[str[i + p][j] - 'A'] = 1;
					dp2[i][j]++;//第j列(i向下推进）从dp2[i][j]向 下没有重复的个数 
				}
			//	printf("# %d -> %d %d\n", i, j, dp2[i][j]);
			}
		}
		int ans = 0;
		for(int i = 0; i < n; ++i) {
			for(int j = 0; j < m; ++j) {
				for(int k = 0; k < dp2[i][j]; ++k) {
					int cnt = INF;
					for(int p = 0; p <= k; ++p) {
						cnt = min(cnt, dp1[i + p][j]);
					}
					int res = k + 1;
					for(int p = 0; p < cnt; ++p) {
						res = min(res, dp2[i][j + p]);
					}
				//	printf("# %d %d\n", cnt, res);
					ans = max(ans, cnt * res);
				}
			}
		}
		printf("%d\n", ans);
	}
    return 0;
}

这道题我看了快一个小时了，还不太理解，先放着吧。