题目链接：http://codeforces.com/contest/682/problem/D
思路：dp[i][j][l][0]表示a串前i和b串前j利用a[i] == b[j]所得到的最长子序列，
dp[i][j][l][1]表示a串前i和b串前j不利用a[i] == b[j]所得到的最长子序列，
所以，dp[i][j][l][0] = max(dp[i-1][j-1][l][0] ,max(dp[i-1][j-1][l-1][0],dp[i-1][j-1][l-1][1])) + 1
dp[i][j][l][1] = max(max(dp[i-1][j][l][0] ,dp[i-1][j][l][1]),max(dp[i][j-1][l][0] ,dp[i][j-1][l][1]))
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e3 + 3;
char a[N],b[N];
int n,m,k;
int dp[N][N][11][2];
int main()
{
scanf("%d %d %d",&n,&m,&k);
scanf("%s %s",a+1,b+1);
for(int i = 1 ;i <= n ;i++)
{
for(int j = 1 ;j <= m ;j++)
{
for(int l = 1 ;l <= k ;l++)
{
if(a[i] == b[j])
dp[i][j][l][0] = max(dp[i-1][j-1][l][0] ,max(dp[i-1][j-1][l-1][0],dp[i-1][j-1][l-1][1])) + 1;
dp[i][j][l][1] = max(max(dp[i-1][j][l][0] ,dp[i-1][j][l][1]),max(dp[i][j-1][l][0] ,dp[i][j-1][l][1]));
}
}
}
printf("%d\n",max(dp[n][m][k][0] ,dp[n][m][k][1]));
return 0;
}