POJ 3614 Sunscreen(贪心+优先队列)
| Time Limit: 1000MS |
|
Memory Limit: 65536K |
| Total Submissions: 5713 |
|
Accepted: 1994 |
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
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：woc，USACO的牛生活真是丰富多彩啊(；′⌒`) 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。
代码如下：
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
struct node1
{
int minspf,maxspf;
}cow[2510];
struct node2
{
int spf,num;
}lotion[2510];
int cmp1(node1 a,node1 b)
{
return a.minspf<b.minspf;
}
int cmp2(node2 a,node2 b)
{
return a.spf<b.spf;
}
int main()
{
int n,m,i,j;
while(scanf("%d%d",&n,&m)!=EOF)
{
for(i=0;i<n;++i)
scanf("%d%d",&cow[i].minspf,&cow[i].maxspf);
for(i=0;i<m;++i)
scanf("%d%d",&lotion[i].spf,&lotion[i].num);
sort(cow,cow+n,cmp1);
sort(lotion,lotion+m,cmp2);
priority_queue<int, vector<int>,greater<int> >q;//维护被选中奶牛的maxspf
j=0;
int ans=0;
for(i=0;i<m;++i)
{
while(j<n&&cow[j].minspf<=lotion[i].spf)
{
q.push(cow[j].maxspf);
j++;
}
while(!q.empty()&&lotion[i].num)
{
int cnt=q.top();
q.pop();
if(cnt>=lotion[i].spf)
{
ans++;
lotion[i].num--;
}
}
}
printf("%d\n",ans);
}
return 0;
}