FZU-2218 Simple String Problem
Problem 2218 Simple String Problem
Accept: 254 Submit: 594
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
第六屆福建省大學生程序設計競賽-重現賽（感謝承辦方華僑大學）
Submit Back Status Discuss 原題地址： 題意： 給你一個串和兩個整數n和k，n表示串的長度，k表示串只有前k個小寫字母，問你兩個不含相同元素的連續子串的長度的最大乘積。 思路: 狀態壓縮DP最多16位，第i位的狀態表示第i位字母是否存在， 代碼:
#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