poj 3614 Sunscreen(优先队列+贪心)

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

Source

USACO 2007 November Gold

 

题意：

有C个奶牛去晒太阳 (1 <=C <= 2500)，每个奶牛各自能够忍受的阳光强度有一个最小值和一个最大值，太大就晒伤了，太小奶牛没感觉。

而刚开始的阳光的强度非常大，奶牛都承受不住，然后奶牛就得涂抹防晒霜，防晒霜的作用是让阳光照在身上的阳光强度固定为某个值。

那么为了不让奶牛烫伤，又不会没有效果。

给出了L种防晒霜。每种的固定的阳光强度和数量也给出来了

每个奶牛只能抹一瓶防晒霜，最后问能够享受晒太阳的奶牛有几个。

 

思路：刚开始想错了，正确的方法为：

          1、将牛按最小值从小到大排好序

          2、将防嗮霜按阳光强度从小到大排好序

          3、然后有一个优先队列，优先队列里按牛的最大值从小到大排序

          4、遍历防晒霜，将牛的最小值<=防晒霜的值的牛push进优先队列，然后根据防晒霜的数量从队首开始选出符合的牛，这部分见代码

          5、while(k<n && cow[k].l<=p[i].x)注意这里的判断条件要写在一起，刚开始分开写，将第二个条件写在下面的if中，结果TLE了。。。。。。

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
#include<vector>
using namespace std;
#define N 2600
int n,m;
struct Node{
    int l,r;
    bool friend operator < (Node a,Node b){
        return a.r>b.r;
    } 
}cow[N];
struct Node1{
    int x,num;
}p[N];
bool cmp(Node a,Node b){
    return a.l<b.l;    
}
bool cmp1(Node1 a,Node1 b){
    return a.x<b.x;
}
int main()
{
    while(scanf("%d%d",&n,&m)==2){
        
        for(int i=0;i<n;i++){
            scanf("%d%d",&cow[i].l,&cow[i].r);
        }
        for(int i=0;i<m;i++){
            scanf("%d%d",&p[i].x,&p[i].num);
        }
        sort(cow,cow+n,cmp);
        sort(p,p+m,cmp1);
        
        //for(int i=0;i<v.size();i++){
        //    printf("---%d
",v[i]);
        //}
        
        priority_queue<Node>q;
        
        int k=0;
        int ans=0;
        for(int i=0;i<m;i++){
            while(k<n && cow[k].l<=p[i].x){
                    q.push(cow[k]);
                    k++;
            }
            
            int w=0;
            while(!q.empty()){
                
                if(w>=p[i].num) break;
                
                Node tmp=q.top();
                if(tmp.r>=p[i].x){
                    q.pop();
                    ans++;
                    w++;
                }
                else{
                    q.pop();
                }
                
            }
        }
        printf("%d
",ans);
    }
    return 0;
}