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
A copy of AC codes:
#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
int cmp(pair<int ,int > a,pair<int ,int >b){
if(a.first==b.first) return a.second<b.second;
else return a.first<b.first;
}
int main(){
int C,L;
cin>>C>>L;
pair<int ,int > c[2510],va[2510];
for(int i=0;i<C;i++)
scanf("%d%d",&c[i].first,&c[i].second);
for(int j=0;j<L;j++)
scanf("%d%d",&va[j].first,&va[j].second);
priority_queue<int ,vector<int >,greater<int > >q;
sort(c,c+C,cmp);
sort(va,va+L,cmp);
int k=0,sum=0;
for(int i=0; i<L; 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;
}
Ideas: The two arrays of data from small to large sort,
Sunscreen enumerate from the smallest, the minimum value of all degrees or less in line with the sun cows placed in the sunscreen maximum priority queue. Then the priority queue is a small value to a minimum so it can be taken out of these maxima. Update the answer.
The whole idea: Sunscreen (small to large) must be supplied to the first maximum value of the minimum cows, other sunscreen to prevent future idle;