POJ 3614 Sunscreen

Sunscreen

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 12907 Accepted: 4534

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 ≤ minSPF_i ≤ 1,000; minSPF_i ≤ maxSPF_i ≤ 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 SPF_i (1 ≤ SPF_i ≤ 1,000). Lotion bottle i can cover cover_i 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: minSPF_i and maxSPF_i
* Lines C+2..C+L+1: Line i+C+1 describes a sunscreen lotion bottle i with space-separated integers: SPF_i and cover_i

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

题解：现在有一群奶牛在沙滩上晒太阳，需要涂防晒霜。求所能被防晒的奶牛最多的只数。

输入：1st 两个用空格分开的整数 C L,分别代表奶牛的头数（cows）和防晒霜的瓶数( lotion )。

2nd 接下来 C行，输入单只奶牛所需最小的防晒霜的防嗮系数 minSPF ,最大的 maxSPF,每行数据用空格隔开。

3rd 再接下来 L行，输入防晒霜的防晒系数t1 和瓶数 t2 每行数据用空格隔开。

注意：不同行的防晒霜，他们的防晒系数可能一样，所以可以用类似于桶排序中的桶的使用方法用一个数组来存储防晒霜瓶数，数组下标就是防晒系数。

思路：读入数据->以奶牛的最小防晒系数为关键字由高到低排列奶牛数据所在的结构体数组->循环遍历奶牛：一只奶牛再遍历最高防晒系数j=maxspf 到最低防晒系数j>=minspf j-- 查询（类似于桶排序）防晒霜数组 lo[j],一旦有防晒霜在

这个区间，那么可被防晒的奶牛数量加一，对应的 lo[j] 即防晒系数spf=j 的防晒霜数量减一。条件是这个防晒霜数量不能少于1.

（所以就是贪心）

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <cstring>
 4 
 5 #include <algorithm>
 6 using namespace std;
 7 //bool vis[2509];
 8 struct cow{
 9     int mina;
10     int maxa;
11 }co[2509];
12 
13 bool cmpcow(cow x, cow y)
14 {
15         return x.mina>y.mina;
16 }
17 int lo[1009];
18 //bool lotions(lotion x,lotion y)
19 //{
20 //    return x.spf<y.spf;
21 //} 
22 int main()
23 {    int c,l;
24         while(scanf("%d%d",&c,&l)!=EOF)
25     {
26     
27         int sum=0;
28     //    memset(vis,false,sizeof(vis));
29     
30         for(int i=1;i<=c;i++){
31             scanf("%d%d",&co[i].mina,&co[i].maxa);
32         }
33         for(int i=1;i<=l;i++){
34             int t1,t2;
35             scanf("%d%d",&t1,&t2);//type
36 //            scanf("");//quantity
37 //            lo[t1].spf=t1;
38             lo[t1]+=t2;
39         }
40         
41         sort(co+1,co+c+1,cmpcow);
42     //    sort(lo,lo+l,lotions);
43         
44         for(int i=1;i<=c;i++){
45             for(int j=co[i].maxa;j>=co[i].mina;j--){
46                 if(lo[j])
47                 {        
48                     sum++,lo[j]--;
49                     break;
50                 }
51             }
52         }
53         
54         printf("%d",sum);    
55     }
56     return 0;
57 }