codeforces 682 D. Alyona and Strings

D. Alyona and Strings

题意

思路

code

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

#define MAXN (1000 + 5)

int n, m, k;
string s, t;
int c[MAXN][MAXN];
int dp[MAXN][MAXN][10];

int max(int a, int b, int c) {
return max(max(a, b), c);
}

int main () {
ios::sync_with_stdio(false);
cin >> n >> m >> k >> s >> t;

memset(c, 0, sizeof(c));
memset(dp, 0, sizeof(dp));
for (int i=0; i<n; i++) {
if (s[i] == t[0]) c[i][0] = 1;
}
for (int j=0; j<m; j++) {
if (s[0] == t[j]) c[0][j] = 1;
}
for (int i=1; i<n; i++) {
for (int j=1; j<m; j++) {
if (s[i] == t[j]) c[i][j] = c[i-1][j-1] + 1;
else c[i][j] = 0;
}
}

for (int i=0; i<n; i++) {
if ((i && dp[i-1][0][1]) || s[i] == t[0])
dp[i][0][1] = 1;
}

for (int i=0; i<m; i++) {
if ((i && dp[0][i-1][1]) || s[0] == t[i])
dp[0][i][1] = 1;
}

for (int v=1; v<=k; v++) {
for (int i=1; i<n; i++) {
for (int j=1; j<m; j++) {
if (c[i][j] != 0) {
int& _c = c[i][j];
if (i - _c >= 0 && j - _c >= 0)
dp[i][j][v] = dp[i - _c][j - _c][v-1] + c[i][j];
else dp[i][j][v] = c[i][j];
}
dp[i][j][v] = max(dp[i][j][v], dp[i-1][j][v], dp[i][j-1][v]);
}
}
}

int ans = 0;
for (int i=1; i<=k; i++) {
ans = max(ans, dp[n-1][m-1][i]);
}
cout << ans << endl;
return 0;
}