POJ 3614-Sunscreen

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

USACO 2007 November Gold
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分析

只用排序然后直接贪心
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程序：

#include<iostream>
using namespace std;
int minspf[2501],maxspf[2501],spf[2501],cover[2501],c,l1;

void kp1()
{
    int cao;
    for (int i=1;i<=c-1;i++)
    {
        for (int j=i+1;j<=c;j++)
        if (maxspf[j]<maxspf[i]||maxspf[i]==maxspf[j]&&maxspf[j]<=maxspf[i])
        {
            cao=minspf[i];minspf[i]=minspf[j];minspf[j]=cao;
            cao=maxspf[i];maxspf[i]=maxspf[j];maxspf[j]=cao;
        }
    }
}

void kp2(int l,int r)
{
    if (l>=r) return;
    int i=l,j=r,mid=spf[(l+r)/2],cao;
    do
    {
        while (spf[i]<mid) i++;
        while (spf[j]>mid) j--;
        if (i<=j)
        {
            cao=spf[i];spf[i]=spf[j];spf[j]=cao;
            cao=cover[i];cover[i]=cover[j];cover[j]=cao;
            i++;j--;
        }
    }while (i<=j);
    kp2(l,j);
    kp2(i,r);
}

int main()
{
    cin>>c>>l1;
    for (int i=1;i<=c;i++)
    cin>>minspf[i]>>maxspf[i];
    for (int i=1;i<=l1;i++)
    cin>>spf[i]>>cover[i];
    kp1();
    kp2(1,l1);
    int tj=0;
    for (int i=1;i<=c;i++)
    {
        for (int j=1;j<=l1;j++)
        {
            if (spf[j]>maxspf[i]) break;
            if (minspf[i]<=spf[j]&&spf[j]<=maxspf[i]&&cover[j])
            {
                tj++;
                cover[j]--;
                break;
            }
        }
    }
    cout<<tj;
    return 0;
}