思路：类似LCS那样的做法只是这里多加了分成K段的要求，那么多加一维就好了
dp[i][j][k][0]表示匹配到s中第i个，t中第j个，组成k段，并且可以向下延续的值
dp[i][j][k][1]表示匹配到s中第i个，t中第j个，组成k段，不可以向下延续的值，转移看代码
#include<bits/stdc++.h>
using namespace std;
int dp[1005][1005][12][2];
char s1[1005],s2[1005];
int main()
{
int n,m,k;
scanf("%d%d%d",&n,&m,&k);
scanf("%s",s1+1);
scanf("%s",s2+1);
for(int i = 1;i<=n;i++)
{
for(int j = 1;j<=m;j++)
{
if(s1[i]==s2[j])
for(int kk = 1;kk<=k;kk++)
dp[i][j][kk][0]=max(dp[i-1][j-1][kk][0],dp[i-1][j-1][kk-1][1])+1;
for(int kk = 1;kk<=k;kk++)
dp[i][j][kk][1]=max(max(dp[i-1][j-1][kk][1],dp[i-1][j][kk][1]),max(dp[i][j-1][kk][1],dp[i][j][kk][0]));
}
}
printf("%d\n",dp[n][m][k][1]);
}
D. Alyona and Strings
time limit per test
memory limit per test
input
output
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.
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.
k non-empty strings p1, p2, p3, ..., pk
- s can be represented as concatenation a1p1a2p2... akpkak + 1, where a1, a2, ..., ak + 1
- t can be represented as concatenation b1p1b2p2... bkpkbk + 1, where b1, b2, ..., bk + 1
- 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.
substring
Input
n, m, k (1 ≤ n, m ≤ 1000, 1 ≤ k ≤ 10) are given — the length of the string s, the length of the string t
s, consisting of lowercase English letters.
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
Note
The following image describes the answer for the second sample case:
