POJ No 3614 Sunscreen 优先队列 贪心

Sunscreen

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 ≤ minSPF_i ≤ 1,000; minSPF_i ≤ maxSPF_i ≤ 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 SPF_i (1 ≤ SPF_i ≤ 1,000). Lotion bottle i can cover cover_i 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: minSPF_i and maxSPF_i 
* Lines C+2..C+L+1: Line i+C+1 describes a sunscreen lotion bottle i with space-separated integers: SPF_i and cover_i

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 <algorithm>
#include <cstring>
#include <cstdio>
#include <queue>
#include <functional>
#include <utility>
using namespace std;

/*
3 2       // C只牛  L个防晒霜
C 行
3 10      minSPF, maxSPF
2 5
1 5
L行
6 2      每种数量, 固定的阳光强度
4 1
*/
const int maxn = 100000;
int C, L;                 // C只奶牛 , L
typedef pair<int, int> P; // minSPF, maxSPF
//小值先出
priority_queue<int, vector<int>, greater<int> > q;
P cow[maxn], bot[maxn];  //牛(min,max) 和 防晒霜(固定的阳光数, 数量) 

void solve();
void input();

void input()
{
    scanf("%d%d", &C, &L);
    for (int i = 0; i < C; i++) {
        scanf("%d%d", &cow[i].first, &cow[i].second);
    }

    for (int i = 0; i < L; i++) {
        scanf("%d%d", &bot[i].first, &bot[i].second);
    }
}

void solve()
{
    input();
    sort(cow, cow + C);    //按照最小值阳光强度升序
    sort(bot, bot + L);    //(first)按照能固定的阳光强度升序
    int j = 0, ans = 0;
    //当  L 个 防晒霜
    for (int i = 0; i < L; i++)
    {
        //将最小值阳光 和 固定的阳光强度 比较
        while (j < C && cow[j].first <= bot[i].first) {
            q.push(cow[j].second);    //添加到优先队列  (最小值阳光)
            j++;
        }        

        //如果 最小值阳光 序列不空，当前 防晒霜未用完
        while (!q.empty() && bot[i].second) {
            int x = q.top(); q.pop();
            // 最小值 小于  能固定的阳光, 方案不存在
            if (x < bot[i].first) continue;
            //否则,
            ans++;
            //当前防晒霜--
            bot[i].second--;
        }
    }
    printf("%d\n", ans);
}

int main()
{
    solve();

    return 0;
}