| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 5538 | Accepted: 1920 |
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
３.，该优先队列维护ｒ的最小值，因为ｒ越大可以适用的护肤品越多（贪心），对压入优先队列的奶牛，如果其ｒ值大于ｉ的值，那么就说明该头奶牛可以使用该护肤品，则立即使用，因为越到后面ｒ值越大，如果它能使用则后面的也肯定可以使用，但是随着ｉ的值的增大，该母牛不能使用的可能性增加，所以应尽快使用（贪心）。并将该奶牛弹出（使用完），如果不能使用，则其ｒ值＜ｉ值，也弹出，因为ｉ只会变大，不会变小。
#include
#include
#include
#include
#include
using namespace std;
struct A
{
int l;
int r;
bool operator< (A a) const{
return r>a.r;
};
}cow[2505];
struct B
{
int f;
int num;
}bot[2505];
bool cmpc(A a,A b)
{
return a.l q;
for(int i=1;i<=n;i++)
{
while(bot[i].f>=cow[u].l)
q.push(cow[u++]);
while(q.size()&&bot[i].num)
{
if(bot[i].f>q.top().r)
{
q.pop();
continue;
}
cnt++;
bot[i].num--;
q.pop();
}
}
printf("%d\n",cnt);
}
return 0;
}