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
优先队列维护，奶牛最小承受值由小到大排序，防晒霜spf值有小到大排序，然后贪心，优先队列维护，最小承受值小的先入队，最大承受能力小的先出队
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
struct cow
{int left,right;
}c[2510];
struct sun
{
int score,num;
}s[2510];
bool cmp1(cow a,cow b)
{
return a.left<b.left;
}
bool cmp2(sun a,sun b)
{
return a.score<b.score;
}
int main()
{
int n,m;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
scanf("%d%d",&c[i].left,&c[i].right);
for(int i=1;i<=m;i++)
scanf("%d%d",&s[i].score,&s[i].num);
sort(c+1,c+1+n,cmp1);
sort(s+1,s+m+1,cmp2);
int j=1;
int ans=0;
priority_queue<int,vector<int>,greater<int> >q;
for(int i=1;i<=m;i++)
{
while(j<=n&&c[j].left<=s[i].score)
{
q.push(c[j].right);
j++;
}
while(!q.empty()&&s[i].num!=0)
{
int w=q.top();
q.pop();
if(w>=s[i].score)
{
ans++;
s[i].num--;
}
}
}
printf("%d\n",ans);
return 0;
}