poj3614 贪心

Sunscreen

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6410		Accepted: 2239

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
题目大意：具体单词我也不太懂，不过大体意思就是每一只奶牛都有着需要的spf最大值和最小值，然后现在有一些类似防晒霜
可以提供spf，如果一只奶牛被保护，意思就是涂的防晒液提供的spf刚好在要求的最大值和最小值之间，问你最多能保护度多
少只奶牛。
思路分析：目前我只会用贪心做，很显然，需要进行排序，但是是按照最大值还是最小值排序呢，用最小值排序很容易举出反例，
按最大值排序的时候很多情况都是符合的，防晒液按照能够提供的spf从小到大排序是毫无疑问的，然后遍历走一遍就能求出最多
的奶牛数了。
讲道理真让我证明我也不会，但是感觉是这样，也许这就是贪心的精髓吧，在没有其他思路的情况下贪心往往能有奇妙的作用。
另外，要细心！！！内层循环j写成了iwa了四发，555555555555.
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
using namespace std;
const int maxn=2500+10;
struct nod
{
    int a;
    int b;
};
nod cow[maxn],s[maxn];
bool cmp1(nod x,nod y)
{
    return x.b==y.b?x.a<y.a:x.b<y.b;
}
bool cmp2(nod x,nod y)
{
    return x.a<y.a;
}
int main()
{
    int c,l;
    int m,n;
    while(scanf("%d%d",&c,&l)!=EOF)
    {
        for(int i=0;i<c;i++)
        {
            scanf("%d%d",&cow[i].a,&cow[i].b);
        }
        for(int i=0;i<l;i++)
        {
            scanf("%d%d",&s[i].a,&s[i].b);
        }
        sort(cow,cow+c,cmp1);
        sort(s,s+l,cmp2);
        int cut=0;
        for(int i=0;i<c;i++)
        {
            for(int j=0;j<l;j++)
            {
                    if(s[j].a>=cow[i].a&&s[j].a<=cow[i].b&&s[j].b>0)
                    {
                        cut++;
                        s[j].b--;
                        break;
                    }
            }
        }
        cout<<cut<<endl;
    }
    return 0;
}