USACO 2007 NOV Sunscreen 防晒霜 贪心

题目

题目描述

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?

有C个奶牛去晒太阳 (1 <=C <= 2500)，每个奶牛各自能够忍受的阳光强度有一个最小值和一个最大值，太大就晒伤了，太小奶牛没感觉。

而刚开始的阳光的强度非常大，奶牛都承受不住，然后奶牛就得涂抹防晒霜，防晒霜的作用是让阳光照在身上的阳光强度固定为某个值。

那么为了不让奶牛烫伤，又不会没有效果。

给出了L种防晒霜。每种的数量和固定的阳光强度也给出来了

每个奶牛只能抹一瓶防晒霜，最后问能够享受晒太阳的奶牛有几个。

输入输出格式

输入格式：

* 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

输出格式：

A single line with an integer that is the maximum number of cows that can be protected while tanning

输入输出样例

输入样例#1： 复制

3 2
3 10
2 5
1 5
6 2
4 1

输出样例#1： 复制

2

分析

中文翻译有点误导，自己看清楚了，防晒霜是先读防晒指数，再读防晒霜的个数。奶牛按上限排序，而防晒霜直接按防晒值排序就可以了。

我对于这种排序方式的理解是：我们使得防晒值与奶牛上限越接近，意味着后面的防晒霜越难来满足这只奶牛，也就是这瓶防晒霜来满足这只奶牛是更优的。一瓶防晒霜最多只能满足一只奶牛，所以如果能满足，那么就用这瓶防晒霜。这就是这道题一个贪心的想法。

程序

#include <bits/stdc++.h>
using namespace std;
const int MAXC = 2500 + 1;
struct node
{
    int Mn, Mx;
}cows[MAXC];
struct sunscreen
{
    int amt, cov;
}ss[MAXC];
bool comp1(node x, node y)
{
    return x.Mx < y.Mx;
}
bool comp2(sunscreen x, sunscreen y)
{
    return x.cov < y.cov;
}
int main()
{
    int c,l,ans;
    cin >> c >> l;
    for (int i = 1; i <= c; i++)
        cin >> cows[i].Mn >> cows[i].Mx;
    for (int i = 1; i <= l; i++)
        cin >> ss[i].cov >> ss[i].amt;
    sort(cows+1, cows+(c+1),comp1);
    sort(ss+1, ss+(l+1), comp2);
    for (int i = 1; i <= c; i++)
    {
        for (int j = 1; j <= l; j++)
            if(ss[j].cov >= cows[i].Mn && ss[j].cov <= cows[i].Mx && ss[j].amt)
            {
                ss[j].amt--;
                ans++;
                break;
            }
    }
    cout << ans << endl;
    return 0;
}