bnuoj 51124 Simple String Problem（状态dp）

Simple String Problem

Time Limit: 2000ms

Memory Limit: 32768KB

This problem will be judged on FZU. Original ID:2218
64-bit integer IO format: %I64d      Java class name:Main

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Recently, you have found your interest in string theory. Here is an interesting question about strings.

You are given a string S of length n consisting of the first k lowercase letters.

You are required to find two non-empty substrings (note that substrings must be consecutive) of S, such that the two substrings don't share any same letter. Here comes the question, what is the maximum product of the two substring lengths?

Input

The first line contains an integer T, meaning the number of the cases. 1 <= T <= 50.

For each test case, the first line consists of two integers n and k. (1 <= n <= 2000, 1 <= k <= 16).

The second line is a string of length n, consisting only the first k lowercase letters in the alphabet. For example, when k = 3, it consists of a, b, and c.

Output

For each test case, output the answer of the question.

Sample Input

425 5abcdeabcdeabcdeabcdeabcde25 5aaaaabbbbbcccccdddddeeeee25 5adcbadcbedbadedcbacbcadbc3 2aaa

Sample Output

6150210

Hint

One possible option for the two chosen substrings for the first sample is "abc" and "de".

The two chosen substrings for the third sample are "ded" and "cbacbca".

In the fourth sample, we can't choose such two non-empty substrings, so the answer is 0.

#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <string>#include <cmath>#include <set>#include <queue>#include <algorithm>#include <vector>const double PI = acos(-1.0);using namespace std;#define esp  1e-8const int inf = 99999999;const int mod = 1000000007;//freopen("in.txt","r",stdin); //输入重定向,输入数据将从in.txt文件中读取//freopen("out.txt","w",stdout); //输出重定向,输出数据将保存在out.txt文件中int dp[70000];char str[2005];int vis[30],vv[20];int main(){int t, n, k, i, j;scanf("%d", &t);while (t--){scanf("%d%d", &n, &k);scanf("%s", str);int len = strlen(str);memset(dp, 0, sizeof(dp));for (i = 0; i < n; ++i){int s = 0;for (j = i; j < n; ++j){s = s | (1 << (str[j] - 'a'));dp[s] = max(dp[s], j - i + 1); //预处理出每种状态的最大值}}int ans = 0;for (i = 0; i < (1 << k); ++i){for (j = 0; j < k; ++j){if (i & (1 << j)){dp[i] = max(dp[i], dp[i ^ (1 << j)]);//字母多的包含字母少的， 更新最大值}}}for (i = 0; i < (1 << k); ++i){ans = max(ans, dp[i] * dp[((1 << k) - 1) ^ i]);}printf("%d\n", ans);}}