标签：esc main span vol mit 分享 长度 recently 题意
Problem 2218 Simple String Problem
Problem DescriptionRecently, 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?
InputThe 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.
OutputFor each test case, output the answer of the question.
Sample Input Sample Output HintOne 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
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<map>
#include<vector>
#include<cstring>
#include<cmath>
#define eps 1e-12
using namespace std;
typedef long long ll;
const ll mo = 1000000007, N = 2*1e3+10;
char s[N];
int dp[(1<<16)+100];
int main()
{
int t;
cin>>t;
while(t--)
{
int n, m;
scanf("%d%d", &n, &m);
scanf("%s", s);
memset(dp, 0, sizeof(dp));
for(int i = 0; i<n; i++)
{
int t = 0;
for(int j = i; j<n; j++)
{
t |= 1<<(s[j] - ‘a‘);
dp[t] = max(dp[t], j - i + 1);
}
}
int s = 1<<m;
for(int i = 0; i<s; i++)
{
for(int j = 0; j<m; j++)
{
if((1<<j) & i)
dp[i] = max(dp[i], dp[i^(1<<j)]);
}
}
int ans = 0;
for(int i = 0; i<s; i++)
{
ans = max(ans, dp[i]*dp[(s-1)^i]);
}
cout<<ans<<endl;
}
return 0;
}
FZU-2218 Simple String Problem
标签：esc main span vol mit 分享 长度 recently 题意
原文地址：https://www.cnblogs.com/luowentao/p/8997048.html
