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...
1 //#pragma comment(linker, "/STACK:102400000, 102400000") 2 #include <algorithm> 3 #include <iostream> 4 #include <cstdlib> 5 #include <cstring> 6 #include <cstdio> 7 #include <vector> 8 #include <cmath> 9 #include <ctime> 10 #include <list> 11 #include <set> 12 #include <map> 13 using namespace std; 14 typedef long long LL; 15 typedef pair <int, int> P; 16 const int N = 1e3 + 5; 17 char str1[N], str2[N]; 18 int dp[N][N][11][2]; 19 20 int main() 21 { 22 int n, m, k; 23 scanf("%d %d %d", &n, &m, &k); 24 scanf("%s %s", str1, str2); 25 for(int i = 1; i <= n; ++i) { 26 for(int j = 1; j <= m; ++j) { 27 for(int s = 1; s <= k; ++s) { 28 if(str1[i - 1] == str2[j - 1]) { 29 dp[i][j][s][0] = max(max(dp[i - 1][j - 1][s - 1][1], dp[i - 1][j - 1][s][0]), 30 dp[i - 1][j - 1][s - 1][0]) + 1; 31 } 32 dp[i][j][s][1] = max(max(dp[i - 1][j][s][1], dp[i][j - 1][s][1]), 33 max(dp[i - 1][j][s][0], dp[i][j - 1][s][0])); 34 dp[i][j][s][1] = max(dp[i][j][s][0], dp[i][j][s][1]); 35 } 36 } 37 } 38 printf("%d\n", dp[n][m][k][1]); 39 return 0; 40 }
