POJ-3614 Sunscreen（优先级队列）

Description

To avoid unsightly burns while tanning, each of the C (1 ≤ C ≤ 2500) cows must cover her hide with sunscreen when they're at the beach. Cow i has a minimum and maximum SPF rating (1 ≤ minSPFi ≤ 1,000; minSPFi ≤ maxSPFi ≤ 1,000) that will work. If the SPF rating is too low, the cow suffers sunburn; if the SPF rating is too high, the cow doesn't tan at all........

The cows have a picnic basket with L (1 ≤ L ≤ 2500) bottles of sunscreen lotion, each bottle i with an SPF rating SPFi (1 ≤ SPFi ≤ 1,000). Lotion bottle i can cover coveri cows with lotion. A cow may lotion from only one bottle.

What is the maximum number of cows that can protect themselves while tanning given the available lotions?

Input

• Line 1: Two space-separated integers: C and L
• Lines 2..C+1: Line i describes cow i's lotion requires with two integers: minSPFi and maxSPFi
• Lines C+2..C+L+1: Line i+C+1 describes a sunscreen lotion bottle i with space-separated integers: SPFi and coveri

Output

A single line with an integer that is the maximum number of cows that can be protected while tanning

Sample Input

3 2
3 10
2 5
1 5
6 2
4 1

Sample Output

2

---

#include <iostream>
#include <vector>
#include <cmath>
#include <string>
#include <climits>
#include <queue>
#include <algorithm>

using namespace std;

const int INF = 0x0ffffff;

struct SPF
{
    int minSPF, maxSPF;
    bool operator<(const SPF & obj) const
    {
        return (minSPF < obj.minSPF || 
            (minSPF == obj.minSPF && maxSPF < obj.maxSPF))
    }
};

struct Node
{
    int value, num;
    bool operator<(const Node & obj) const
    {
        return value < obj.value;
    }
};

int C = 2505, L = 2505;
vector<SPF> cow(C);
vector<Node> bottle(L);

int sunScreen()
{
    sort(cow.begin() + 1, cow.begin() + 1 + C);
    sort(bottle.begin() + 1, bottle.begin() + 1 + L);

    int pos = 1;
    int cnt = 0;
    priority_queue<int, vector<int>, greater<int> > pq;

    for (int i = 1; i <= L; ++i) {
        //所有满足minSPF<=value的牛进入队列
        while (pos <= C && bottle[i].value >= cow[pos].minSPF) {
            pq.push(cow[pos++].maxSPF);
        }

        while (!pq.empty() && bottle[i].num) {
            int tmp = pq.top(); pq.pop();
            if (bottle[i].value > tmp) continue;
            ++cnt;
            --bottle[i].num;
        }
    }

    return cnt;
}

int main()
{
    std::ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

    cin >> C >> L;
    for (int i = 1; i <= C; ++i) cin >> cow[i].minSPF >> cow[i].maxSPF;
    for (int i = 1; i <= L; ++i) cin >> bottle[i].value >> bottle[i].num;
    cout << sunScreen() << endl;

    return 0;
}

每个牛的SPF按左端点（minSPF）排序，防晒霜按照从小到大排序，思路是首先防晒霜的防护范围应该不小于minSPF，然后选择最小的maxSPF，这样后面防晒霜的防护范围增大的时候，后面可选择的范围也就相应增大了。