【LittleXi】Plan a part-time job

【LittleXi】Plan a part-time job

topic link

Mental journey:

I got up in the morning and opened the topic, I didn’t have any ideas. I
clicked on “Dyson Ball” and it was noon. I quit the game, glanced at the line segment tree, and it took a second.
Summary: Playing games can also improve thinking ability

problem solving ideas

Line segment tree + discretization + dp
discretization: because the time is too long (1e9), we should sort the time, and use the order 1, 2, 3 to simulate the time order
dp: sort the line segments by the first node, for each The maximum profit of a node can be transferred from the maximum profit of all previous nodes, that is, nowprofit=the maximum profit of all previous nodes+profit[now] Line segment
tree: When looking for the maximum profit before, we can think of using the line segment tree to find the interval Maximum
Time Complexity: O(nlogn)

the code

#define lc k<<1
#define rc k<<1|1

const int maxn = 500010;
const int inf = 0x3f3f3f3f;
int n, a[maxn];

struct node
{
    
    
    int l, r, mx;
};
node tree[maxn * 4];

void build(int k, int l, int r)//创建叶子节点，k表示存储下标，l,r表示更新区间
{
    
    
    tree[k].l = l;
    tree[k].r = r;
    if (l == r)
    {
    
    
        tree[k].mx = a[l];
        return;
    }
    int mid = (l + r) / 2;
    build(lc, l, mid);
    build(rc, mid + 1, r);
    tree[k].mx = max(tree[lc].mx, tree[rc].mx);
}

void update(int k, int i, int v)//点更新，将a[i]修改为v
{
    
    
    if (tree[k].l == tree[k].r && tree[k].l == i)
    {
    
    
        tree[k].mx = v;
        return;
    }
    int mid = (tree[k].l + tree[k].r) / 2;
    if (i <= mid)
        update(lc, i, v);
    else
        update(rc, i, v);
    tree[k].mx = max(tree[lc].mx, tree[rc].mx);
}

int query(int k, int l, int r)//区间覆盖查询，求区间[l,r]的最大值
{
    
    
    if (tree[k].l >= l && tree[k].r <= r)
        return tree[k].mx;
    int mid = (tree[k].l + tree[k].r) / 2;
    int ma = -inf;
    if (l <= mid)
        ma = max(ma, query(lc, l, r));
    if (r > mid)
        ma = max(ma, query(rc, l, r));
    return ma;
}


class Solution {
    
    
public:
    int jobScheduling(vector<int>& st, vector<int>& en, vector<int>& pr) {
    
    
        int n = st.size();
        vector<vector<int>> vec(n);
        set<int> lisan;
        for (int i = 0; i < n; i++)
        {
    
    
            lisan.insert(st[i]);
            lisan.insert(en[i]);
        }
        int p = 1;
        unordered_map<int, int> m;
        for (auto& val:lisan)
        {
    
    
            m[val] = p++;
        }
        for (int i = 0; i < n; i++)
        {
    
    
            vec[i].push_back(m[st[i]]);
            vec[i].push_back(m[en[i]]);
            vec[i].push_back(pr[i]);
        }
        sort(vec.begin(), vec.end(), [&](vector<int>& v1,
            vector<int>& v2) {
    
    return v1[0] < v2[0]; });
        memset(a, 0, sizeof(a));
        build(1, 0, 500005);
        for (int i = 0; i < n; i++)
        {
    
    
            int be = vec[i][0];
            int en = vec[i][1];
            int money = vec[i][2];
            int temp = query(1, 1, be);
            a[en] = max(a[en],temp + money);
            update(1, en, a[en]);
        }
        int re = *max_element(a, a + 500000);
        return re;
    }
};