Sunscreen
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 8435 | Accepted: 2981 |
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
题意：有C头奶牛要去沐光浴，太阳光太强烈会晒坏皮肤，太弱又会没效果。每头牛都有一个太阳光适宜的范围经行沐光浴，分别给出minspf_i和maxspf_i。 有L种防晒霜，每种防晒霜可以把所受阳光固定于一个值spf_i，每种有cover_i瓶。 问最多会有几头牛得到合适的光晒强度？
题解：
贪心策略，在满足minspf的条件下，尽量将spf的防晒霜涂到maxspf小的奶牛身上，因为maxspf大的奶牛有更多的选择。
这里就需要一个优先队列来储存满足minspf的奶牛的maxspf的值。 具体解题步骤如下：
1.将奶牛按照minspf升序排列，将防晒霜按照spf升序排列。
2.枚举防晒霜，将minspf<=spf的奶牛的maxspf存到优先队列中，然后值小的先出队列，
看是否满足maxspf>=spf，更新记录值。
#include<bits/stdc++.h>
using namespace std;
struct cow
{
int min_spf,max_spf;
};
struct suns
{
int spf,num;
};
vector<cow> c;
vector<suns> s;
int n,m;
bool cmp1(cow a,cow b)
{
return a.min_spf<b.min_spf;
}
bool cmp2(suns a,suns b)
{
return a.spf<b.spf;
}
priority_queue<int,vector<int>,greater<int> >qua;
int main()
{
cin>>n>>m;
for(int i=0;i<n;i++)
{
cow tmp;
cin>>tmp.min_spf>>tmp.max_spf;
c.push_back(tmp);
}
for(int i=0;i<m;i++)
{
suns tmp;
cin>>tmp.spf>>tmp.num;
s.push_back(tmp);
}
sort(c.begin(),c.end(),cmp1);
sort(s.begin(),s.end(),cmp2);
int ans=0,cnt=0;
for(int i=0;i<m;i++)
{
while(cnt<n&&c[cnt].min_spf<=s[i].spf){
qua.push(c[cnt].max_spf);
cnt++;
}
while(s[i].num!=0&&!qua.empty()){
int k=qua.top();
qua.pop();
if(k>=s[i].spf)
{
ans++;s[i].num--;
}
}
}
cout<<ans<<endl;
}