牛客算法周周练3

链接：https://ac.nowcoder.com/acm/contest/5338/E
来源：牛客网

时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld

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?

输入描述:

* 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

输出描述:

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

示例1

输入

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

输出

2

题意：

有c只牛，l种防晒液，每种牛有一个舒适范围，防晒液能够让牛位于某个值SPF，但有数量限制，问最多能让几头牛处在舒适范围内

题解：

这题我们贪心就好了，我们将奶牛的最小SPFi按照从小到大排序，同时将防晒乳液的SPFi从小到大排序

那么我们遍历防晒液，对于每种防晒液，我们将之前的最小SPFi值比这种防晒乳液SPFi值小的牛加到优先队列里，然后每次我们优先让最大SPFi小的奶牛使用当前这种防晒乳液。

#include <bits/stdc++.h>
typedef long long LL;
#define pb push_back
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
const int mod = 1e9+7;
const int maxn = 1e5+10;
using namespace std;

struct cow
{
    int l,r;
}A[3005];
struct bo
{
    int val;
    int num;
}B[3005];
bool cmp1(cow a, cow b)
{
    if(a.l!=b.l) return a.l<b.l;
    else return a.r<b.r;
}
bool cmp2(bo a, bo b)
{
    if(a.val!=b.val) return a.val<b.val;
    else return a.num<b.num;
}
priority_queue<int ,vector<int>,greater<int> > qe;

int main()
{
    #ifdef DEBUG
    freopen("sample.txt","r",stdin); //freopen("data.out", "w", stdout);
    #endif
    
    int n,m;
    scanf("%d %d",&n,&m);
    for(int i=1;i<=n;i++)
        scanf("%d %d",&A[i].l,&A[i].r);
    for(int i=1;i<=m;i++)
        scanf("%d %d",&B[i].val,&B[i].num);
    sort(A+1,A+1+n,cmp1);
    sort(B+1,B+1+m,cmp2);
    int ans = 0;
    int p = 1;
    for(int i=1;i<=m;i++)
    {
        while(p<=n && A[p].l<=B[i].val)//将l比这种防晒液val小的牛的r入队
        {
            qe.push(A[p].r);
            p++;
        }
        while(!qe.empty() && B[i].num>0)//优先让r小的奶牛使用
        {
            int t = qe.top(); qe.pop();
            if(t >= B[i].val)//之后的防晒液val值只会更大,所以r<val的出队也不影响后面
            {
                ans++;
                B[i].num--;
            }
        }
    }
    printf("%d
",ans);
    
    return 0;
}

 另一种贪心写法：

#include <bits/stdc++.h>
typedef long long LL;
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
const int mod = 1e9+7;
const int maxn = 1e5+10;
using namespace std;

vector<pair<int,int> > vt;
map<int,int> mp;

int main()
{
    #ifdef DEBUG
    freopen("sample.txt","r",stdin); //freopen("data.out", "w", stdout);
    #endif
    
    int n,m;
    scanf("%d %d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        int l,r;
        scanf("%d %d",&l,&r);
        vt.push_back({l,r});
    }
    for(int i=1;i<=m;i++)
    {
        int val,num;
        scanf("%d %d",&val,&num);
        mp[val]+=num;
    }
    sort(vt.begin(),vt.end());
    mp[0] += 1;
    int ans = 0;
    for(int i=n-1;~i;i--)
    {
        map<int,int>::iterator it = mp.upper_bound(vt[i].second);
        it--;
        if(it->first >= vt[i].first)
        {
            ans++;
            it->second--;
            if(it->second == 0) mp.erase(it);
        }
    }
    printf("%d
",ans);
    
    return 0;
}

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