FZU 2218 Simple String Problem (状压dp)

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Simple String Problem

Time Limit: 2000 mSec Memory Limit : 32768 KB

Problem Description

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

4
25 5
abcdeabcdeabcdeabcdeabcde
25 5
aaaaabbbbbcccccdddddeeeee
25 5
adcbadcbedbadedcbacbcadbc
3 2
aaa

Sample Output

6
150
21
0

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.

Source

第六届福建省大学生程序设计竞赛-重现赛（感谢承办方华侨大学）

题目链接：http://acm.fzu.edu.cn/problem.php?pid=2218

题目大意：给一个长度为n由k种字母组成的字符串，求其中两个子串，要求它们不包含相同的字母且长度的乘积最大

题目分析：很明显的状态压缩了，dp[sta]表示选字母的状态为sta时的字符串最大长度，这个可以n^2得到，先考虑最后的结果，对于某个状态sta符合条件的可以与之相乘的状态易知为(2^k - 1) 异或 sta，做到这里已经可以过样例了，但并不完全正确，还需要枚举每一种状态的子集更新某个状态下的最大值，比如说aaacccabc，包含bc的状态为110，假设我现在选了a也就是001的状态，那可以与之相乘的状态包括110，100，010，我需要取它们的最大值赋给110，由于字母可以重复，因此完全可能出现子集的长度更大，上面的例子中110即bc对应的长度为2，而100即ccc却是3，因此我们要将110状态下的值修改为3

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int const CON = 16;
int dp[(1 << CON) + 5];
char s[2005];

int main() {
	int T;
	scanf("%d", &T);
	while(T --) {
		memset(dp, 0, sizeof(dp));
		int n, k, sta, tot;
		scanf("%d %d", &n, &k);
		scanf("%s", s);
		tot = 1 << k;
		for(int i = 0; i < n; i++) {
			sta = 0;
			for(int j = i; j < n; j++) {
				sta |= (1 << (s[j] - 'a'));
				dp[sta] = max(dp[sta], j - i + 1);
			}
		}
		for(int i = 0; i < tot; i++) {
			for(int j = 0; j < k; j++) {
				if(i & (1 << j)) {
					dp[i] = max(dp[i], dp[i - (1 << j)]);
				}
			}
		}
		int ans = 0;
		for(int i = 0; i < tot; i++) {
			ans = max(ans, dp[i] * dp[(tot - 1) ^ i]);
		}
		printf("%d\n", ans);
	}
}