动态规划基础（三）最长上升子序列LIS

这个有三个板子，分别是$dp$做法，二分+贪心做法和树状数组优化$dp$，树状数组这个我后面学了之后再补上哈

题目描述

给定$n$个元素，要求找到最长上升子序列的长度

$dp$做法

#include

using namespace std;

int a[1005], p[1005];

int main()
{
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int n;cin >> n;
    for (int i = 1;i <= n;i++) {
        cin >> a[i];
        p[i] = 1;
    }
    for (int i = 1;i <= n;i++) 
        for (int j = 1;j <= i;j++) 
            if (a[j] < a[i])p[i] = max(p[j] + 1, p[i]);
    int mi = 0;
    for (int i = 1;i <= n;i++)mi = max(mi, p[i]);
    cout << mi;
    return 0;
}

二分+贪心做法

#include

using namespace std;

int a[1005];
int dp[1005];

int main()
{
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int n;cin >> n;
    int len = 1;
    for (int i = 1;i <= n;i++) {
        cin >> a[i];
        dp[i] = INT_MIN;
    }
    dp[1] = a[1];
    for (int i = 2;i <= n;i++) {
        if (a[i] > dp[len])dp[++len] = a[i];
        else {
            int pl = lower_bound(dp + 1, dp + 1 + len, a[i]) - dp;
            dp[pl] = a[i];
        }
    }
    cout << len;
    return 0;
}