Codeforces 682 D. Alyona and Strings (dp)

题目链接：http://codeforces.com/contest/682/problem/D

给你两个字符串，求两个字符串中顺序k个的相同子串 长度之和。(注意是子串)

dp[i][j][k][0] 表示a[i] == a[j]时，a字符串前i个和b字符串前j个，顺序k个相同的子串 长度之和

dp[i][j][k][1] 表示a[i] != a[j]时，顺序k个相同子串的长度之和

dp[i][j][k][0] = max(dp[i - 1][j - 1][k][0], dp[i - 1][j - 1][k - 1][1], dp[i - 1][j - 1][k - 1]) + 1;

dp[i][j][k][1] = max(dp[i - 1][j][k][1], dp[i][j - 1][k][1], dp[i][j][k][0], dp[i - 1][j][k][0], dp[i][j - 1][k][0]);

渣代码...

//#pragma comment(linker, "/STACK:102400000, 102400000")
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
typedef pair <int, int> P;
const int N = 1e3 + 5;
char str1[N], str2[N];
int dp[N][N][11][2];

int main()
{
    int n, m, k;
    scanf("%d %d %d", &n, &m, &k);
    scanf("%s %s", str1, str2);
    for(int i = 1; i <= n; ++i) {
        for(int j = 1; j <= m; ++j) {
            for(int s = 1; s <= k; ++s) {
                if(str1[i - 1] == str2[j - 1]) {
                    dp[i][j][s][0] = max(max(dp[i - 1][j - 1][s - 1][1], dp[i - 1][j - 1][s][0]), 
                        dp[i - 1][j - 1][s - 1][0]) + 1;
                }
                dp[i][j][s][1] = max(max(dp[i - 1][j][s][1], dp[i][j - 1][s][1]), 
                    max(dp[i - 1][j][s][0], dp[i][j - 1][s][0]));
                dp[i][j][s][1] = max(dp[i][j][s][0], dp[i][j][s][1]);
            }
        }
    }
    printf("%d
", dp[n][m][k][1]);
    return 0;
}