Codeforces Round #595 (Div. 3) D2. Too Many Segments (hard version) [线段树]

D2. Too Many Segments (hard version)

https://codeforces.com/contest/1249/problem/D2
题意 ：给了n个线段区间让你尽可能的删最少的线段 使每个区间线段覆盖次数少于k次
思维题 D1直接贪过去 我们找这个区间最远R的 剪掉 这样一定对后面的最优
D2 暴力就找最远的显然不合适了
这里我想的 优先队列 存之前到这个点所有的线段 用Vector 存L点 开始出现的线段
之后 我们直接开始从1 遍历到 N 那个点大于K 我们选择最远的R 来把它-1 直到小于等于K

#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
const int N = 2e5;
int n, k;
struct Seg {
    int id;
    int l, r;
    bool operator < (const Seg & a) const {
        return r < a.r;
    }
} seg[maxn];
vector<Seg> G[maxn];
priority_queue<Seg> que;
vector<int> ans;

int tree[maxn << 2], add[maxn << 2];
inline void push_up(int rt) {
    tree[rt] = max(tree[rt << 1], tree[rt << 1 | 1]);
}

inline void push_down(int rt) {
    if(!add[rt])
        return ;
    tree[rt << 1] += add[rt];
    tree[rt << 1 | 1] += add[rt];
    add[rt << 1] += add[rt];
    add[rt << 1 | 1] += add[rt];
    add[rt] = 0;
}

void updata(int L, int R, int l, int r, int rt, int val) {
    //cout << l << " " << r << endl;
    if(L <= l && R >= r) {
        tree[rt] += val;
        add[rt] += val;
        return ;
    }
    push_down(rt);
    int mid = l + r >> 1;
    if(L <= mid)
        updata(L, R, l, mid, rt << 1, val);
    if(R > mid)
        updata(L, R, mid + 1, r, rt << 1 | 1, val);
    push_up(rt);
}

int query(int L, int l, int r, int rt) {
    if(l == r)
        return tree[rt];
    push_down(rt);
    int mid = l + r >> 1;
    if(L <= mid)
        return query(L, l, mid, rt << 1);
    else
        return query(L, mid + 1, r, rt << 1 | 1);
}

int main() {
    cin >> n >> k;
    for(int i = 1; i <= n; i ++) {
        cin >> seg[i].l >> seg[i].r;
        seg[i].id = i;
        G[seg[i].l].push_back(seg[i]);
        updata(seg[i].l, seg[i].r, 1, N, 1, 1);
    }

// for(int i = 1; i <= 15; i ++) {
// cout << query(i, 1, N, 1) << " ";
// } cout << endl;

// sort(seg + 1, seg + 1 + n, [](const Seg & a, const Seg & b) {
// return a.l < b.l || (a.l == b.l && a.r < b.r);
// }); // 本来想遍历线段 发现不用 复习下这个写法 lambda 函数 匿名函数

    for(int i = 1; i <= N; i ++) {
        for(Seg t : G[i]) que.push(t);
        while(query(i, 1, N, 1) > k) {
            Seg t = que.top();
            que.pop();
            ans.push_back(t.id);
            updata(t.l, t.r, 1, N, 1, -1);
        }
    }
    cout << ans.size() << endl;
    for(int i = 0; i < ans.size(); i ++) {
        if(i != 0) cout << " ";
        cout << ans[i];
        if(i == ans.size() - 1) cout << endl;
    }
    return 0;
}