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 ≤ 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 
 2 #include 
 3 #include 
 4 
 5 #include 
 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.spf21 //} 
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 }

 

。。。

转载于:https://www.cnblogs.com/greenaway07/p/10662213.html

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