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  • 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:

     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;
    }
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  • 原文地址:https://www.cnblogs.com/wsy107316/p/11392718.html
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