Coffee Break

Posted 2022-08-13 wsy107316

题目链接：Coffee Break  Gym-101911A

题目大意：有一位员工想要利用喝咖啡来休息，他给了一个数组表示他想要喝咖啡的时间点（假设他喝咖啡用时1分钟），老板规定每次喝咖啡的时间间隔必须要大于d。问：他将给定数组的时间点都经

　　　　　　历一遍最少（贪心所在）需要多长时间，并输出每个时间点是在第几天经历的；

解题思路：贪心地选取喝咖啡的时间，我们尽量选取喝咖啡时间靠前的，然后贪心寻找能放在他后面最靠前的时间点，如果放不开就新开一天。

用map和队列的做法：

 

 1 /* */
 2 # include <iostream>
 3 # include <stdio.h>
 4 # include <string.h>
 5 # include <string>
 6 # include <iomanip>
 7 # include <algorithm>
 8 # include <ctime>
 9 # include <cmath>
10 # include <climits>
11 # include <cstdlib>
12 # include <utility>
13 # include <bitset>
14 # include <cctype>
15 # include <cassert>
16 # include <set>
17 # include <map>
18 # include <deque>
19 # include <queue>
20 # include <stack>
21 # include <vector>
22 # include <functional>
23 using namespace std;
24 
25 typedef long long ll;
26 const int maxn=2e5+50;
27 const ll mod=1e9+7;
28 const int eps=1e-8;
29 const double pi=acos(-1.0);
30 # define mem(a,x) memset((a),(x),sizeof((a)))
31 # define gcd(a,b) (__gcd(a, b))
32 # define lcm(a,b) (a*b/__gcd(a, b))
33 # define lson l,m,rt<<1
34 # define rson m+1,r,rt<<1|1
35 # define lowbit(x)(x&(-x))
36 
37 int origin[maxn], after[maxn];
38 map<int, int>m;///存某时间喝咖啡是在哪一天
39 //priority_queue<int, vector<int>, greater<int> > q;
40 queue<int>q;
41 int main()
42 
43     int n, mm, d, cnt=1;
44     cin>>n>>mm>>d;
45 
46     for(int i=1; i<=n; i++ )
47     
48         scanf("%d", &origin[i]);
49         after[i] = origin[i];
50     
51 
52     sort(after+1, after+1+n);
53     m[after[1]] = 1;//时间点最靠前的在第一天
54     q.push(1);
55     for(int i=2; i<=n; i++ )
56     
57         int top=q.front();
58         if( after[i]-after[top]>d )//可以在同一天
59         
60             m[after[i]]=m[after[top]];
61             q.pop();
62         
63 
64         else
65         
66             cnt++;
67             m[after[i]] = cnt;
68         
69 
70         q.push(i);
71     
72     cout<<cnt<<endl;
73     for(int i=1; i<=n; i++ )
74     
75         if( i==1 )
76             printf("%d", m[origin[i]]);
77 
78         else
79             printf(" %d", m[origin[i]]);
80     
81     cout<<endl;
82     return 0;
83 

 

set的做法：

/* */
# include <iostream>
# include <stdio.h>
# include <string.h>
# include <string>
# include <iomanip>
# include <algorithm>
# include <ctime>
# include <cmath>
# include <climits>
# include <cstdlib>
# include <utility>
# include <bitset>
# include <cctype>
# include <cassert>
# include <set>
# include <map>
# include <deque>
# include <queue>
# include <stack>
# include <vector>
# include <functional>
using namespace std;

typedef long long ll;
const int maxn=2e5+50;
const ll mod=1e9+7;
const int eps=1e-8;
const double pi=acos(-1.0);
# define mem(a,x) memset((a),(x),sizeof((a)))
# define gcd(a,b) (__gcd(a, b))
# define lcm(a,b) (a*b/__gcd(a, b))
# define lson l,m,rt<<1
# define rson m+1,r,rt<<1|1
# define lowbit(x)(x&(-x))

int a[maxn];
set<int>s;
map<int,int>mp;//喝咖啡的时间点在第几天
int n, m, d;
int main()

    int i;
    cin>>n>>m>>d;
    s.clear();//清空set
    for( i=1; i<=n; i++ )
    
        cin>>a[i];
        s.insert(a[i]);//插入时间点
    

    set<int>::iterator iter;//迭代器指针iter
    int cnt=0;
    int ans=1;

    while( s.size() )//当set容器中的元素个数不为0
    
        iter = s.lower_bound(cnt);//在set中查找第一个大于等于cnt的数的所在位置,如果不在返回s.end()
        if( iter==s.end())
        
            ans++;
            cnt=0;
        

        else
        
            mp[*iter] = ans;
            s.erase(*iter);//删除set中值为*iter的数
            cnt = *iter+d+1;
        
    
    cout<<ans<<endl;
    for(int i=1; i<=n; i++ )
    
        if( i==1 )
            printf("%d", mp[a[i]]);
        else
            printf(" %d", mp[a[i]]);
    
    cout<<endl;
    return 0;