FOJ Problem 2218 Simple String Problem

最新推荐文章于 2025-08-17 13:50:03 发布

这个做不了。

最新推荐文章于 2025-08-17 13:50:03 发布

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本文探讨了一个关于字符串理论的问题，要求从给定的字符串中找到两个不包含相同字母的子串，并求得这两个子串长度的最大乘积。介绍了输入输出格式及样例，并提供了AC代码实现。

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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

AC代码：

# include <stdio.h>
# include <string.h> 
# include <algorithm>
using namespace std;
int w[20], dp[1<<17];
char s[2010];
int main(){
	int i, j, k, n, t, l, status;
	for(i=0; i<16; i++)
	w[i]=(1<<i);
	scanf("%d", &t);
	while(t--){
		memset(dp, 0, sizeof(dp));
		scanf("%d%d", &l, &k);
		getchar();
		scanf("%s", s);
		for(i=0; i<l; i++){
			status=0;
			for(j=i; j<l; j++){
				status=(status|w[s[j]-'a']);
				dp[status]=max(dp[status], j-i+1);
			}
		}
		int end=(1<<k)-1;
		for(i=0; i<=end; i++){
			for(j=0; j<k; j++){
				if(i&w[j])
				dp[i]=max(dp[i], dp[i^w[j]]);
			}
		}
		int Max=0;
		for(i=1; i<end; i++)
		Max=max(dp[i]*dp[(1<<k)-1-i], Max);  
		printf("%d\n", Max);
	}
	return 0;
}