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

输出

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。

 1 #include <bits/stdc++.h>
 2 typedef long long LL;
 3 #define pb push_back
 4 const int INF = 0x3f3f3f3f;
 5 const double eps = 1e-8;
 6 const int mod = 1e9+7;
 7 const int maxn = 1e5+10;
 8 using namespace std;
 9 
10 struct cow
11 {
12     int l,r;
13 }A[3005];
14 struct bo
15 {
16     int val;
17     int num;
18 }B[3005];
19 bool cmp1(cow a, cow b)
20 {
21     if(a.l!=b.l) return a.l<b.l;
22     else return a.r<b.r;
23 }
24 bool cmp2(bo a, bo b)
25 {
26     if(a.val!=b.val) return a.val<b.val;
27     else return a.num<b.num;
28 }
29 priority_queue<int ,vector<int>,greater<int> > qe;
30 
31 int main()
32 {
33     #ifdef DEBUG
34     freopen("sample.txt","r",stdin); //freopen("data.out", "w", stdout);
35     #endif
36     
37     int n,m;
38     scanf("%d %d",&n,&m);
39     for(int i=1;i<=n;i++)
40         scanf("%d %d",&A[i].l,&A[i].r);
41     for(int i=1;i<=m;i++)
42         scanf("%d %d",&B[i].val,&B[i].num);
43     sort(A+1,A+1+n,cmp1);
44     sort(B+1,B+1+m,cmp2);
45     int ans = 0;
46     int p = 1;
47     for(int i=1;i<=m;i++)
48     {
49         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
50         {
51             qe.push(A[p].r);
52             p++;
53         }
54         while(!qe.empty() && B[i].num>0)//优先让r小的奶牛使用
55         {
56             int t = qe.top(); qe.pop();
57             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
58             {
59                 ans++;
60                 B[i].num--;
61             }
62         }
63     }
64     printf("%d
",ans);
65     
66     return 0;
67 }

另一种贪心写法：

 1 #include <bits/stdc++.h>
 2 typedef long long LL;
 3 const int INF = 0x3f3f3f3f;
 4 const double eps = 1e-8;
 5 const int mod = 1e9+7;
 6 const int maxn = 1e5+10;
 7 using namespace std;
 8 
 9 vector<pair<int,int> > vt;
10 map<int,int> mp;
11 
12 int main()
13 {
14     #ifdef DEBUG
15     freopen("sample.txt","r",stdin); //freopen("data.out", "w", stdout);
16     #endif
17     
18     int n,m;
19     scanf("%d %d",&n,&m);
20     for(int i=1;i<=n;i++)
21     {
22         int l,r;
23         scanf("%d %d",&l,&r);
24         vt.push_back({l,r});
25     }
26     for(int i=1;i<=m;i++)
27     {
28         int val,num;
29         scanf("%d %d",&val,&num);
30         mp[val]+=num;
31     }
32     sort(vt.begin(),vt.end());
33     mp[0] += 1;
34     int ans = 0;
35     for(int i=n-1;~i;i--)
36     {
37         map<int,int>::iterator it = mp.upper_bound(vt[i].second);
38         it--;
39         if(it->first >= vt[i].first)
40         {
41             ans++;
42             it->second--;
43             if(it->second == 0) mp.erase(it);
44         }
45     }
46     printf("%d
",ans);
47     
48     return 0;
49 }