2021.7.14 刷题记录

AcWing 1012友好城市

#include <iostream>
#include <cstdio>
#include <algorithm>

using namespace std;

const int N = 5e3+5;

int n;

struct Node{
    int a, b;
    bool operator<(const Node& t) const{
        return a < t.a;
    }
}node[N];

int d[N], len = 0; //d[i]:长度为i的最长上升子序列的最后一位的最小值(递增)

int main()
{
    scanf("%d", &n);
    for(int i = 0; i < n; ++ i)
    {
        scanf("%d%d", &node[i].a, &node[i].b);
    }
    sort(node, node+n);
    for(int i = 0; i < n; ++ i)
    {
        //找到大于等于(等号取不到)node[i].b的最小值，替换他，如果为len+1，表示添加到结尾
        int l = 1, r = len+1;
        while(l < r)
        {
            int mid = (l+r)>>1;
            if(d[mid] >= node[i].b) //(等号其实取不到)
            {
                r = mid;
            }
            else
            {
                l = mid+1;
            }
        }
        d[l] = node[i].b;
        len = max(len, l);
    }
    printf("%d\n", len);
    return 0;
}

AcWing1016. 最大上升子序列和

#include <iostream>
#include <cstdio>

using namespace std;

const int N = 1e5+5;

int n;
int nums[N];

int dp[N]; //dp[i]表示以第i位结尾的最大上升子序列的和

int main()
{
    scanf("%d", &n);
    for(int i = 1; i <= n; ++ i)
    {
        scanf("%d", nums+i);
    }
    for(int i = 1; i <= n; ++ i)
    {
        dp[i] = nums[i];
        for(int j = 1; j < i; ++ j)
        {
            if(nums[j] < nums[i])
            {
                dp[i] = max(dp[i], dp[j]+nums[i]);
            }
        }
    }
    int ans = 0;
    for(int i = 1; i <= n; ++ i)
    {
        ans = max(ans, dp[i]);
    }
    printf("%d\n", ans);
    return 0;
}