Codeforces Round #358 (Div. 2) D. Alyona and Strings

D. Alyona and Strings

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

After returned from forest, Alyona started reading a book. She noticed strings s and t, lengths of which are n and m respectively. As usual, reading bored Alyona and she decided to pay her attention to strings s and t, which she considered very similar.

Alyona has her favourite positive integer k and because she is too small, k does not exceed 10. The girl wants now to choose k disjoint non-empty substrings of string s such that these strings appear as disjoint substrings of string t and in the same order as they do in string s. She is also interested in that their length is maximum possible among all variants.

Formally, Alyona wants to find a sequence of k non-empty strings p1, p2, p3, …, pk satisfying following conditions:

s can be represented as concatenation a1p1a2p2... akpkak + 1, where a1, a2, ..., ak + 1 is a sequence of arbitrary strings (some of them may be possibly empty);
t can be represented as concatenation b1p1b2p2... bkpkbk + 1, where b1, b2, ..., bk + 1 is a sequence of arbitrary strings (some of them may be possibly empty);
sum of the lengths of strings in sequence is maximum possible.

Please help Alyona solve this complicated problem and find at least the sum of the lengths of the strings in a desired sequence.

A substring of a string is a subsequence of consecutive characters of the string.

Input

In the first line of the input three integers n, m, k (1 ≤ n, m ≤ 1000, 1 ≤ k ≤ 10) are given — the length of the string s, the length of the string t and Alyona’s favourite number respectively.

The second line of the input contains string s, consisting of lowercase English letters.

The third line of the input contains string t, consisting of lowercase English letters.

Output

In the only line print the only non-negative integer — the sum of the lengths of the strings in a desired sequence.

It is guaranteed, that at least one desired sequence exists.

Examples

Input

3 2 2

abc

ab

Output

2

Input

9 12 4

bbaaababb

abbbabbaaaba

Output

7

题意

给两个字符串s,t，s和t中最多各取k段，并且s和t对应每段都相同，问k段长度之和最大为多少。

思路

dp，记dp1[i][j][x]为到s[i],t[j],目前取了x段，且s[i],t[j]是第x段的结尾时，答案最大为多少，dp2[i][j][x]为到s[i],t[j],目前取了x段，且s[i]或t[j]不是第x段段的结尾时，答案最大为多少。

转移时，如果s[i]==t[j]，则

dp1[i][j][x]=max(dp1[i-1][j-1][x],dp2[i-1][j-1][x-1])+1;
//这里需要注意，dp1[i-1][j-1][x]不为0或x=1时才可转移

否则dp1[i][j][x]=0

dp2[i][j][x]转移时为dp1[i-1][j][x]，dp1[i][j-1][x]，dp2[i-1][j][x]，dp2[i][j-1][x]中最大的一个

然后在题目所给范围内，穷举i,j,k的情况，最终结果为x取1到k，dp1
[m][x]，dp2
[m][x]中最大的。时间复杂度o(nmk)

代码

#include <cstdio>
#include <algorithm>
using namespace std;
char s[1005],t[1005];
int dp1[1005][1005][11],dp2[1005][1005][11];
int main()
{
int n,m,k,ans=0;
scanf("%d %d %d",&n,&m,&k);
scanf("%s",s);
scanf("%s",t);
for (int i=1;i<=n;i++)
{
for (int j=1;j<=m;j++)
{
for (int x=1;x<=k;x++)
{
dp2[i][j][x]=max(max(dp1[i-1][j][x],dp1[i][j-1][x]),max(dp2[i-1][j][x],dp2[i][j-1][x]));
if (s[i-1]==t[j-1])
{
if (dp1[i-1][j-1][x] || dp2[i-1][j-1][x-1] || x==1)
{
dp1[i][j][x]=max(dp1[i-1][j-1][x],dp2[i-1][j-1][x-1])+1;
}
}
}
}
}
for (int x=1;x<=k;x++) ans=max(ans,max(dp1
[m][x],dp2
[m][x]));
printf("%d",ans);
}

PS

知道是要DP，想不出状态。。看了题解之后又感觉好简单，前文写的解题思路基本翻译了一遍cf上官方的英文题解。。感觉需要训练一下做DP类的题的解题思路。