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 + ;
 char str1[N], str2[N];
 int dp[N][N][][];

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