Coffee Break

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：

 /* */
 # 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+;
 const ll mod=1e9+;
 const int eps=1e-;
 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<<
 # define rson m+,r,rt<<|
 # define lowbit(x)(x&(-x))
 
 int origin[maxn], after[maxn];
 map<int, int>m;///存某时间喝咖啡是在哪一天
 //priority_queue<int, vector<int>, greater<int> > q;
 queue<int>q;
 int main()
 {
     int n, mm, d, cnt=;
     cin>>n>>mm>>d;
 
     for(int i=; i<=n; i++ )
     {
         scanf("%d", &origin[i]);
         after[i] = origin[i];
     }
 
     sort(after+, after++n);
     m[after[]] = ;//时间点最靠前的在第一天
     q.push();
     for(int i=; i<=n; i++ )
     {
         int top=q.front();
         if( after[i]-after[top]>d )//可以在同一天
         {
             m[after[i]]=m[after[top]];
             q.pop();
         }
 
         else
         {
             cnt++;
             m[after[i]] = cnt;
         }
 
         q.push(i);
     }
     cout<<cnt<<endl;
     for(int i=; i<=n; i++ )
     {
         if( i== )
             printf("%d", m[origin[i]]);
 
         else
             printf(" %d", m[origin[i]]);
     }
     cout<<endl;
     return ;
 }

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+;
const ll mod=1e9+;
const int eps=1e-;
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<<
# define rson m+,r,rt<<|
# 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=; 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=;
    int ans=;
 
    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=;
        }
 
        else
        {
            mp[*iter] = ans;
            s.erase(*iter);//删除set中值为*iter的数
            cnt = *iter+d+;
        }
    }
    cout<<ans<<endl;
    for(int i=; i<=n; i++ )
    {
        if( i== )
            printf("%d", mp[a[i]]);
        else
            printf(" %d", mp[a[i]]);
    }
    cout<<endl;
    return ;
}