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

题目链接：

题目

D. Alyona and Strings
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output

问题描述

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.

输入

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.

输出

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.

样例

input
9 12 4
bbaaababb
abbbabbaaaba

output
7

题意

求由两个字符串的k个公共子串按顺序拼成的最长子序列。

题解

dp[i][j][kk][0]表示串s1[0...i]和s2[0...j]的kk个公共子串能拼的最长长度，并且最后一个字串还能继续连下去。
dp[i][j][kk][1]表示串s1[0...i]和s2[0...j]的kk个公共子串能拼的最长长度，并且最后一个字串不能继续连下去。

代码

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;

const int maxn = 1111;
typedef __int64 LL;

LL 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%s", s1+1, s2+1);
	memset(dp, 0, sizeof(dp));
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= m; j++) {
			for (int kk = 1; kk <= k; kk++) {
				if (s1[i] == s2[j]) {
					dp[i][j][kk][0] = max(dp[i - 1][j - 1][kk][0], dp[i - 1][j - 1][kk - 1][1]) + 1;
				}
				dp[i][j][kk][1] = max(dp[i - 1][j][kk][1], dp[i][j - 1][kk][1]);
				dp[i][j][kk][1] = max(dp[i][j][kk][1], dp[i][j][kk][0]);
			}
		}
	}
	printf("%I64d\n", dp[n][m][k][1]);
	return 0;
}