POJ 3614 Sunscreen （优先队列）

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

 

样例解释：第一种防晒霜的固定阳光强度6可以用于第一头 [3，10] 的牛，答案加一，第一种防晒霜两瓶变成一瓶；

第二种防晒霜的固定阳光强度4可以用于第二头 [2，5] 的牛或者第三头 [1，5] 的牛，但是这种防晒霜只有一瓶，

所以答案加一，第二种防晒霜没了，结束，所以答案是2。

思路：看懂样例的话，理解题意应该就没问题了。这题应该想到用贪心的思想，对奶牛的阳光承受区间从小到大排序，

对防晒霜的固定阳光强度从小到大排序，然后用优先队列处理就好了......

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<set>
#include<vector>
#include<queue>
using namespace std;
#define ll long long 
const int inf=1e9+7;
const int mod=1e9+7;

const int maxn=3e3+10;

bool cmp(const pair<int,int> &a,const pair<int,int> &b)
{
    if(a.first != b.first)//优先排min_spfi 
        return a.first < b.first;
    else if(a.first == b.first)//min_spfi相同排max_spfi 
        return a.second < b.second;
}

int main()
{
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    
    priority_queue<int,vector<int>,greater<int> >q;//优先出队小的元素 
    
    /*q.push(1);q.push(5);q.push(3);//test 
    
    while(!q.empty())
    {
        cout<<q.top()<<endl;
        q.pop();
    }*/
    
    pair<int,int> cow[maxn],you[maxn];//定义两个pair数组，比较方便 
    
    int c,l;
    cin>>c>>l;
    
    for(int i=0;i<c;i++)//输入每只奶牛的min_spfi和max_spfi 
        cin>>cow[i].first>>cow[i].second;
    
    for(int i=0;i<l;i++)//输入每瓶油（防晒霜2333）的固定阳光强度和瓶数 
        cin>>you[i].first>>you[i].second;
        
    sort(cow,cow+c,cmp);//排序 
    
    sort(you,you+l,cmp);//排序 
    
    //for(int i=0;i<c;i++)//test 
    //    cout<<cow[i].first<<" "<<cow[i].second<<endl;
    
    ll int ans=0;//记录答案 
    
    int now=0;
    
    for(int i=0;i<l;i++)//对每瓶油对奶牛遍历 
    {
        while(now<c&&cow[now].first<=you[i].first)//固定阳光强度大于奶牛的min_spfi 
        {
            q.push(cow[now].second);//入队 
            now++;//找下一头奶牛 
        }
        
        while(!q.empty())//队列不为空 
        {
            if(you[i].second==0)//油没了直接break 
                break;
            
            int x=q.top();//找出奶牛的max_spfi 
            q.pop();
            
            if(x>=you[i].first)//固定阳光强度不超过牛的max_spfi 
            {
                ans++;//成功 
                you[i].second--;//油减一瓶 
            }
            else//否则跳过 
                continue;
    
        }
        
    }
    
    cout<<ans<<endl;//输出答案 
        
    return 0;
}