poj3614Sunscreen

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

现附上AC代码：

#include
#include
#include
#include
using namespace std;
int cmp(pair a,pairb){
if(a.first==b.first) return a.second else return a.first}
int main(){
int C,L;
cin>>C>>L;
pair c[2510],va[2510];
for(int i=0;i scanf("%d%d",&c[i].first,&c[i].second);
for(int j=0;j scanf("%d%d",&va[j].first,&va[j].second);
priority_queue,greater >q;
sort(c,c+C,cmp);
sort(va,va+L,cmp);
int k=0,sum=0;
for(int i=0; i while(k < C && c[k].first <= va[i].first){
q.push(c[k].second); k++;
}
while( !q.empty() && va[i].second)
{
int m = q.top();
q.pop();
if(m >= va[i].first) sum++,va[i].second--;
}
}
printf("%d",sum);
return 0;
}

 

思路：将两个数组的数据从小到大进行排序，

从最小的防晒霜枚举，将所有符合 防晒度的最小值小于等于该防晒霜的奶牛 最大值放入优先队列之中。然后优先队列是小值先出所以就可以将这些最大值中的最小的取出来。更新答案。

整体思路：防晒霜（由小到大）肯定是最先供给给最大值最小的奶牛，防止以后其他防晒霜闲置；