Sunscreen
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 10937 | Accepted: 3835 |
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 1Sample Output
2Source
思路：
把牛的数组，minSPF值从小到大排，相等情况下，maxSPF值大的优先，然后，把防晒霜的值也按从小到大排序。
- 对于每一瓶防晒，把minSPF小于等于防晒SPF的牛的maxSPF值push进优先队列，对于这瓶防晒，只能在push进的这些牛里选（没push进的，其minSPF大于防晒霜SPF不满足牛的要求）
- 对于每一个在队列中的牛，我们优先选择maxSPF小的，（但是这个最小值也要大于等于该防晒霜的SPF值）因为大的有更多的选择机会，这个防晒霜处理完有两种情况（1、队列的牛小于防晒霜数，该队列所有牛都满足，防晒霜剩余。2、防晒霜不够用，队列还剩牛，这时候处理下一个防晒霜，下一个防晒霜的spf肯定比这个防晒霜大，因此在队列的牛都满足minSPF<=nextSPF）
其实我也证明不出来这个思想就一定正确，但是在没别的方法下，不如用贪心思想，因为它真的hin神奇。
代码如下：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
typedef pair<int,int> P;
P cow[2505],sun[2505];
int main(){
int c,l;
scanf("%d%d",&c,&l);
for(int i=0;i<c;i++){
scanf("%d%d",&cow[i].first,&cow[i].second);
}
for(int i=0;i<l;i++){
scanf("%d%d",&sun[i].first,&sun[i].second);
}
sort(cow,cow+c);
sort(sun,sun+l);
priority_queue<int,vector<int>,greater<int> > qq;
int j=0,ans=0;
for(int i=0;i<l;i++){
while(j<c&&cow[j].first<=sun[i].first){
qq.push(cow[j].second);
j++;
}
while(!qq.empty()&&sun[i].second){
int mymin=qq.top();
qq.pop();
if(mymin>=sun[i].first){
ans++;
sun[i].second--;
}
}
}
printf("%d\n",ans);
}