POJ-3614-Sunscreen

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?

输入：

  • 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：

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

输入示例：

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

输出示例：

1
2

题目大意：

有C头牛，每个牛Ci在一个给定的区间上需要涂防晒霜，有L种防晒霜，给出每个防晒霜适合的数值SPF[i]和个数cover[i]，求能最多满足牛的个数。

贪心策略：

按区间的开始位置递减排序，依次考虑每头牛。
对于每头牛，选取满足在此区间内最大的防晒霜。

Solution1（暴力版）：

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#include<iostream>
#include<utility>
#include<cstdio>
#include<map>
#include<algorithm>
using namespace std;
typedef pair<int,int> PII;
const int MAXN = 5000+10;
int C,L;
PII p[MAXN];
PII q[MAXN];
int main(){
    int ans = 0;
    scanf("%d%d",&C,&L);
    for(int i = 0; i < C; i++){
        scanf("%d%d",&p[i].first,&p[i].second);
    }
    sort(p,p+C);
    for(int i = 0; i < L; i++){
        int a,b;
        scanf("%d%d",&a,&b);
        q[i].first = a;
        q[i].second = b;
    }
    sort(q,q+L);
	//从开始位置大到小考察每头牛。
    for(int i = C-1; i >= 0; i--){
		//同样，对于每头牛，考察在该牛区间内最右的防晒霜。
        for(int j = L-1; j >= 0; j--){
            if(q[j].second > 0 && q[j].first >= p[i].first && q[j].first <= p[i].second){
                ans++;
                q[j].second--;
                break;
            }
        }
    }
    printf("%d",ans);
    return 0;
}

Solution2（平衡树版）：

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#include<iostream>
#include<cstdio>
#include<map>
#include<algorithm>
using namespace std;
const int MAXN = 2510;
typedef pair<int,int> PII;
int N,M;
PII cows[MAXN];
int main(){
    cin >> N >>M;
    for(int i = 0; i < N; i++){
        cin >> cows[i].first >> cows[i].second;
    }
    sort(cows,cows+N);
    map<int,int> spfa;
    for(int i = 0; i < M; i++){
        int spa,cover;
        cin >> spa >> cover;
        spfa[spa] += cover;
    }
    spfa[0] = spfa[1001] = N;
    int ans = 0;
    for(int i = N-1; i >= 0; i--){
        map<int,int>::iterator it = spfa.upper_bound(cows[i].second);
        --it;
        if(cows[i].first <= it->first && cows[i].second >= it->first){
            ans++;
            if(-- it->second == 0) spfa.erase(it);
        }
    }
    cout << ans;
    return 0;
}