pat1017. Queueing at Bank (25)

1017. Queueing at Bank (25)

时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue

Suppose a bank has K windows open for service. There is a yellow line in front of the windows which devides the waiting area into two parts. All the customers have to wait in line behind the yellow line, until it is his/her turn to be served and there is a window available. It is assumed that no window can be occupied by a single customer for more than 1 hour.

Now given the arriving time T and the processing time P of each customer, you are supposed to tell the average waiting time of all the customers.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 numbers: N (<=10000) - the total number of customers, and K (<=100) - the number of windows. Then N lines follow, each contains 2 times: HH:MM:SS - the arriving time, and P - the processing time in minutes of a customer. Here HH is in the range [00, 23], MM and SS are both in [00, 59]. It is assumed that no two customers arrives at the same time.

Notice that the bank opens from 08:00 to 17:00. Anyone arrives early will have to wait in line till 08:00, and anyone comes too late (at or after 17:00:01) will not be served nor counted into the average.

Output Specification:

For each test case, print in one line the average waiting time of all the customers, in minutes and accurate up to 1 decimal place.

Sample Input: 
7 3
07:55:00 16
17:00:01 2
07:59:59 15
08:01:00 60
08:00:00 30
08:00:02 2
08:03:00 10
Sample Output: 
8.2

提交代码

 

堆的常见操作：

  1 #include<set>
  2 #include<map>
  3 #include<cstdio>
  4 #include<algorithm>
  5 #include<iostream>
  6 #include<cstring>
  7 #include<queue>
  8 #include<vector>
  9 #include<cmath>
 10 using namespace std;
 11 #define open 28800
 12 #define close 61200
 13 struct custom{
 14     int come,cost,finish;
 15 };
 16 void swap(custom &a,custom &b){
 17     custom c=a;
 18     a=b;
 19     b=c;
 20 }
 21 void BuildHeap(custom *cc,int m){
 22     int fa,child=m-1,i;
 23     for(i=(child-1)/2;i>=0;i--){
 24         child=i*2+1;//左儿子 
 25         for(fa=i;child<m;child=fa*2+1){
 26             if(child+1<m&&cc[child].finish>cc[child+1].finish){
 27                 child++;
 28             }
 29             if(cc[child].finish<cc[fa].finish){
 30                 swap(cc[fa],cc[child]); 
 31                 fa=child;
 32             }
 33             else{
 34                 break;
 35             }
 36         }
 37     } 
 38 }
 39 void Insertion(custom *cc,custom cur,int &m){
 40     int i=m++;
 41     for(;i>0&&cc[(i-1)/2].finish>cur.finish;i=(i-1)/2){
 42         cc[i]=cc[(i-1)/2];
 43     }
 44     cc[i]=cur;
 45 }
 46 custom DeleteMin(custom *cc,int &m){
 47     custom cur=cc[0];
 48     custom temp=cc[--m];
 49     int fa,child=1;
 50     for(fa=0;child<m;child=fa*2+1){
 51         if(child<m-1&&cc[child].finish>cc[child+1].finish){
 52             child++;
 53         }
 54         if(cc[child].finish<temp.finish){
 55             cc[fa]=cc[child];
 56             fa=child;//保证fa指向当前要比较的节点 
 57         }
 58         else{
 59             break;
 60         }
 61     } 
 62     cc[fa]=temp;
 63     return cur;
 64 }
 65 bool cmp(custom a,custom b){
 66     return a.come<b.come;
 67 }
 68 int main(){
 69     //freopen("D:\\input.txt","r",stdin);
 70     int n,nn;
 71     int i,j;
 72     scanf("%d %d",&n,&nn);
 73     
 74     //cout<<n<<" "<<nn<<endl;
 75     
 76     custom *c=new custom[n+5],*cc=new custom[nn+5];
 77     int h,m,s,cost;
 78     for(i=0;i<n;i++){
 79         scanf("%d:%d:%d %d",&h,&m,&s,&cost);
 80         c[i].come=h*3600+m*60+s;
 81         c[i].cost=cost*60;
 82     }
 83     int totaltime=0,count=0;
 84     sort(c,c+n,cmp);
 85     
 86     
 87     j=1;
 88     for(i=0;i<nn&&i<n;i++){
 89         if(c[i].come<open){
 90             totaltime+=open-c[i].come;
 91             c[i].come=open;
 92         }
 93         if(c[i].come>close){
 94             break;
 95         }
 96         c[i].finish=c[i].come+c[i].cost;
 97         cc[i]=c[i];
 98         count++;
 99     }
100     
101     //cout<<count<<endl;
102     
103     if(count<nn){//人数不够 
104         printf("%.1lf\n",totaltime*1.0/60/count);//不经意间看到，让我找了将近一小时！！ 
105         return 0;
106     }
107 
108     BuildHeap(cc,count);//建堆
109     
110     
111     custom cur;
112     for(;i<n;i++){
113         cur=DeleteMin(cc,nn);
114         if(c[i].come<=close){//cur.finish<=close&&
115             if(cur.finish<c[i].come){
116                 c[i].finish=c[i].come+c[i].cost;
117             }
118             else{
119                 c[i].finish=cur.finish+c[i].cost;
120                 totaltime+=cur.finish-c[i].come;
121             }
122             cur=c[i];
123             Insertion(cc,cur,nn);
124             count++;
125         }
126         else{
127             break;
128         }
129     }
130     printf("%.1lf\n",totaltime*1.0/count/60);
131     return 0;
132 }

 

posted @ 2015-08-28 17:45  Deribs4  阅读(283)  评论(0)  编辑  收藏  举报