D. Alyona and Strings
题目连接：
http://www.codeforces.com/contest/682/problem/D
Description
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.
Sample Input
3 2 2
abc
ab
Sample Output
2
Hint
题意
给你两个串，s和t，现在要你在s中找k个不相交的子串，然后在t中这k个都出现，而且顺序和在s中一致，让你最大化长度和
题解：
从数据范围一看，应该是一个nmk的dp
dp[i][j][k][0]表示s中第i个等于t中第j个，一共匹配了k段后，且继续往下延续，这时候的总长度为多少。
dp[i][j][k][1]表示匹配了k段后，不往下延续的答案是多少
然后像LCS一样转移就好了。
代码
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1005;
int dp[maxn][maxn][11][2];
char s1[maxn],s2[maxn];
int n,m,k;
int main()
{
scanf("%d%d%d",&n,&m,&k);
scanf("%s",s1+1);
scanf("%s",s2+1);
int ans = 0;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
if(s1[i]==s2[j])
for(int t=1;t<=k;t++)
dp[i][j][t][0]=max(dp[i-1][j-1][t][0],dp[i-1][j-1][t-1][1])+1;
for(int t=1;t<=k;t++)
dp[i][j][t][1]=max({dp[i-1][j-1][t][1],dp[i-1][j][t][1],dp[i][j-1][t][1],dp[i][j][t][0]});
}
}
cout<<dp[n][m][k][1]<<endl;
}