洛谷 P3045 [USACO12FEB]Cow Coupons G

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#include <bits/stdc++.h>
using namespace std;
using PII = pair<int, int>;
const int N = 50000;
int n, k, ans;
struct Node {
    int p;
    int c;
} cow[N];
priority_queue<int, vector<int>, greater<int>> dq; // 折扣小根堆
priority_queue<PII, vector<PII>, greater<PII>> cq, pq; // [折后价,坐标]小根堆，[原价,坐标]小根堆
long long m, cost;
bool st[N];
int main() {
    scanf("%d%d%lld", &n, &k, &m);
    for (int i = 0; i < n; ++i) scanf("%d%d", &cow[i].p, &cow[i].c);
    sort(cow, cow + n, [](Node a, Node b) {
        return a.c < b.c; // 按折后价排序
    });
    for (int i = 0; i < k; ++i) { // 尽可能将 k 个优惠券用掉
        cost += cow[i].c;
        if (cost > m) {
            printf("%d\n", i);
            return 0;
        }
        dq.push(cow[i].p - cow[i].c);
    }
    if (k == n) {
        printf("%d\n", n);
        return 0;
    }
    for (int i = k; i < n; ++i) {
        cq.emplace(cow[i].c, i);
        pq.emplace(cow[i].p, i);
    }
    int ans = k;
    // 将剩下的钱花掉，要保持每一笔花费尽可能小，以此买更多的牛
    for (int i = k; i < n; ++i) {
        // 已买过的不再考虑购买
        while (st[cq.top().second]) cq.pop();
        while (st[pq.top().second]) pq.pop();
        auto [c, ci] = cq.top();
        auto [p, pi] = pq.top();
        // 要么用优惠券买 cow[ci]，花费 c，还需要原价购买 dp.top() 对应的牛
        int t1 = c + dq.top();
        // 要么买价格最小的，花费 p
        int t2 = p;
        if (t1 < t2) { // 前者花费更少
            cost += t1;
            dq.pop();
            dq.push(cow[ci].p - cow[ci].c);
            st[ci] = 1;
        } else { // 后者花费更少
            cost += t2;
            st[pi] = 1;
        }
        ++ans;
        if (cost > m) {
            printf("%d\n", ans - 1);
            return 0;
        }
    }
    printf("%d\n", n);
    return 0;
}