LIS (最长上升子序列)

LIS两种写法

O(n^2)

dp[i]表示以a[i]结尾的为LIS长度

#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 = 2e3 + 5;
int dp[N];
int a[N];

int main()
{
    int n;
    while(~scanf("%d", &n)) {
        memset(dp, 0, sizeof(dp));
        int res = 0;
        for(int i = 1; i <= n; ++i) {
            scanf("%d", a + i);
            dp[i] = 1;
            for(int j = 1; j < i; ++j) {
                if(a[i] > a[j])
                    dp[i] = max(dp[i], dp[j] + 1);
            }
            res = max(res, dp[i]);
        }
        printf("%d
", res);
    }
    return 0;
}

O(nlogn)

dp[i]表示LIS长度为i的最后一个元素

//#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 = 1e5 + 5;
int dp[N], a[N], inf = 1e6;

int main()
{
    int n;
    while(~scanf("%d", &n)) {
        for(int i = 0; i <= n + 1; ++i)
            dp[i] = inf;
        for(int i = 1; i <= n; ++i) {
            scanf("%d", a + i);
            *lower_bound(dp, dp + n, a[i]) = a[i];
        }
        printf("%d
", lower_bound(dp, dp + n, inf) - dp);
    }
    return 0;
}