$O(n^3)$超时
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 3010;
int n;
int a[N], b[N];
int f[N][N];
int main()
{
scanf("%d", &n);
for (int i = 1; i <= n; i ++ ) scanf("%d", &a[i]);
for (int i = 1; i <= n; i ++ ) scanf("%d", &b[i]);
for (int i = 1; i <= n; i ++ )
for (int j = 1; j <= n; j ++ )
{
f[i][j] = f[i - 1][j];
if (a[i] == b[j])
{
f[i][j] = max(f[i][j], 1);
for (int k = 1; k < j; k ++ )
if (b[k] < b[j])
f[i][j] = max(f[i][j], f[i][k] + 1);
}
}
int res = 0;
for (int i = 1; i <= n; i ++ ) res = max(res, f[n][i]);
printf("%d", res);
return 0;
}
$O(n^2)偷懒写法$
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 3010;
int n;
int a[N], b[N];
int f[N][N];
int main()
{
scanf("%d", &n);
for (int i = 1; i <= n; i ++ ) scanf("%d", &a[i]);
for (int i = 1; i <= n; i ++ ) scanf("%d", &b[i]);
for (int i = 1; i <= n; i ++ )
{
int maxv = 1;
for (int j = 1; j <= n; j ++ )
{
f[i][j] = f[i - 1][j];
if (a[i] == b[j])
f[i][j] = max(f[i][j], maxv);
if (b[j] < a[i]) maxv = max(maxv, f[i][j] + 1);
}
}
int res = 0;
for (int i = 1; i <= n; i ++ ) res = max(res, f[n][i]);
printf("%d", res);
return 0;
}
不偷懒写法(超内存)
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 3001;
int n;
int a[N], b[N];
int g[N][N]; // g[i][j]表示满足a[i] > b[j] 的所有 f[i][j] + 1 的最大值
int f[N][N];
int main()
{
scanf("%d", &n);
for (int i = 1; i <= n; i ++ ) scanf("%d", &a[i]);
for (int i = 1; i <= n; i ++ ) scanf("%d", &b[i]);
for (int i = 1; i <= n; i ++ )
{
g[i][0] = 1;
for (int j = 1; j <= n; j ++ )
{
f[i][j] = f[i - 1][j];
if (a[i] == b[j])
f[i][j] = max(f[i][j], g[i][j - 1]);
g[i][j] = g[i][j - 1];
if (b[j] < a[i]) g[i][j] = max(g[i][j], f[i][j] + 1);
}
}
int res = 0;
for (int i = 1; i <= n; i ++ ) res = max(res, f[n][i]);
printf("%d", res);
return 0;
}