PAT_A 1016. Phone Bills (25)

1016. Phone Bills (25)

A long-distance telephone company charges its customers by the following rules:

Making a long-distance call costs a certain amount per minute, depending on the time of day when the call is made. When a customer starts connecting a long-distance call, the time will be recorded, and so will be the time when the customer hangs up the phone. Every calendar month, a bill is sent to the customer for each minute called (at a rate determined by the time of day). Your job is to prepare the bills for each month, given a set of phone call records.

Input Specification:

Each input file contains one test case. Each case has two parts: the rate structure, and the phone call records.

The rate structure consists of a line with 24 non-negative integers denoting the toll (cents/minute) from 00:00 - 01:00, the toll from 01:00 - 02:00, and so on for each hour in the day.

The next line contains a positive number N (<= 1000), followed by N lines of records. Each phone call record consists of the name of the customer (string of up to 20 characters without space), the time and date (mm:dd:hh:mm), and the word “on-line” or “off-line”.

For each test case, all dates will be within a single month. Each “on-line” record is paired with the chronologically next record for the same customer provided it is an “off-line” record. Any “on-line” records that are not paired with an “off-line” record are ignored, as are “off-line” records not paired with an “on-line” record. It is guaranteed that at least one call is well paired in the input. You may assume that no two records for the same customer have the same time. Times are recorded using a 24-hour clock.

Output Specification:

For each test case, you must print a phone bill for each customer.

Bills must be printed in alphabetical order of customers’ names. For each customer, first print in a line the name of the customer and the month of the bill in the format shown by the sample. Then for each time period of a call, print in one line the beginning and ending time and date (dd:hh:mm), the lasting time (in minute) and the charge of the call. The calls must be listed in chronological order. Finally, print the total charge for the month in the format shown by the sample.

Sample Input:

10 10 10 10 10 10 20 20 20 15 15 15 15 15 15 15 20 30 20 15 15 10 10 10

10

CYLL 01:01:06:01 on-line

CYLL 01:28:16:05 off-line

CYJJ 01:01:07:00 off-line

CYLL 01:01:08:03 off-line

CYJJ 01:01:05:59 on-line

aaa 01:01:01:03 on-line

aaa 01:02:00:01 on-line

CYLL 01:28:15:41 on-line

aaa 01:05:02:24 on-line

aaa 01:04:23:59 off-line

Sample Output:

CYJJ 01

01:05:59 01:07:00 61 12.10Totalamount:12.10

CYLL 01

01:06:01 01:08:03 122 24.4028:15:4128:16:05243.85

Total amount: 28.25aaa0102:00:0104:23:594318638.80

Total amount: $638.80

题目分析：

排序（字符串排序，时间排序）

on-off：排序后，使用最新的on，off使用最早的，形成on-off。on1 on2 off1 off2 ,计算on2-off1时间差。

费用计算，不同时间段的费率不同

code:

#include<iostream>
#include<algorithm>
#include<vector>
#include<cstring>
#include<cstdio>
using namespace std;
vector<double> rate;
double rate_sum=0;
struct _com
{
string name;
int mm;
int dd;
int hh;
int mm2;
bool isOn;
};
/*
* compare of string && cstring
* 这里说的字母排序仅仅是按照前列字母的大小排序的，前列相等时，
* 接着拿取后一个字母
* */
int comp(_com a,_com b)
{
if(a.name==b.name)
{
if(a.mm<b.mm)
return true;
else if(a.mm>b.mm)
return false;

if(a.dd<b.dd)
return true;
else if(a.dd>b.dd)
return false;

if(a.hh<b.hh)
return true;
else if(a.hh>b.hh)
return false;

if(a.mm2<b.mm2)
return true;
else if(a.mm2>b.mm2)
return false;

return false;
}else
return a.name<b.name;
}
vector<_com> com;

int main()
{
int tmp_rate=0;
for(int i=0;i<24;i++)
{
cin>>tmp_rate;
rate.push_back(tmp_rate);
rate_sum+=tmp_rate;
}
int tmp_count=0;
cin>>tmp_count;
_com tmp_com;
string tmp_line;
for(int i=0;i<tmp_count;i++)
{
cin>>tmp_com.name;
scanf("%2d:%2d:%2d:%2d",&tmp_com.mm,
&tmp_com.dd,
&tmp_com.hh,
&tmp_com.mm2);
cin>>tmp_line;
if(strcmp(tmp_line.c_str(),"on-line")==false)
tmp_com.isOn=true;
else
tmp_com.isOn=false;
com.push_back(tmp_com);
}
sort(com.begin(),com.end(),comp);

int count=1;
//这一步其实，没必要，不再修改了
_com com_tmp=com.at(0);
for(int i=0;i<com.size();i++)
{
if(com_tmp.name!=com.at(i).name)
{
com_tmp=com.at(i);
count++;
}
}
com_tmp=com.at(0);
int j=0;
double total=0;//部分和
int isBegin=0;//总和
int isShow=0;//
for(int i=0;i<count;i++)
{
for(;j<com.size();j++)
{
if(com_tmp.name==com.at(j).name)//&&j<com.size()-1)
{
if(com.at(j).isOn==true)
{
com_tmp=com.at(j);
isBegin=1;

}
//已经成功配对
else if(com.at(j).isOn==false&&isBegin==1)
{
com_tmp=com.at(j-1);
_com com_back=com_tmp;
//on
double tmp=0;
int tmp_count=0;
while(com_back.dd<com.at(j).dd||
com_back.hh<com.at(j).hh||
com_back.mm2<com.at(j).mm2)
{
tmp+=rate.at(com_back.hh);
tmp_count++;
if(com_back.mm2==59)
{
if(com_back.hh==23)
com_back.dd+=1;
com_back.hh=(com_back.hh+1)%24;
}
com_back.mm2=(com_back.mm2+1)%60;
}

/*
* 错误范例，
10 10 10 10 10 10 20 20 20 15 15 15 15 15 15 15 20 30 20 15 15 10 10 10
2
CYLL 01:01:06:01 on-line
CYLL 01:01:06:02 off-line
这样算的就明显错误了。

错误计算方法，以后计算时，尽量减少以分类计算这种计算方法，情况列举很容易丢失
建议：采取更通用的方法，比如更改后的计算方法。之后要加强这方面的训练。
double tmp=(60-com_tmp.mm2)*rate.at(com_tmp.hh)+
com.at(j).mm2*rate.at(com.at(j).hh);
//hh
if(com_tmp.dd!=com.at(j).dd)
{
//不在同一天，分三步
//1
for(int i=com_tmp.hh+1;i<24;i++)
{
tmp+=rate.at(i)*60;
}
//2.day
for(int i=com_tmp.dd+1;i<com.at(j).dd;i++)
{
tmp+=rate_sum*60;
}
//3.
for(int i=0;i<com.at(j).hh;i++)
{
tmp+=rate.at(i)*60;
}
}else
{
//同一天,直接
//1
for(int i=com_tmp.hh+1;i<com.at(j).hh;i++)
{
tmp+=rate.at(i)*60;
}
}
*/
total+=tmp;
if(total>0)
{
if(isShow==0)
{
printf("%s %02d\n",com_tmp.name.c_str(),com_tmp.mm);
isShow=1;
}

printf("%02d:%02d:%02d %02d:%02d:%02d %d $%.2f\n",
com_tmp.dd,com_tmp.hh,com_tmp.mm2,
com.at(j).dd,com.at(j).hh,com.at(j).mm2,
tmp_count,
tmp/100
);

}

isBegin=0;
}
}else
{
com_tmp=com.at(j);
break;
}
}
if(total>0)
printf("Total amount: $%.2f\n",total/100);
total=0;
isShow=0;
isBegin=0;
}
return 0;
}
/*************************************参考别人的代码*****************************************
* 1.比较算法，这个自己要联系
* 2.处理时间差部分,计算费用
* */
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<vector>
#include<map>
using namespace std;
const int HOURS=24;
const int NAME=21;
typedef struct Record{
char name[NAME];
int mou;
int dd;
int hh;
int mm;
char line[10];
}Record;
typedef struct Res{
float money;
int last_time;
}Res;
int compare_name(const void *a ,const void *b){
int tmp=strcmp(((Record*)a)->name,((Record *)b)->name);
if(tmp==0){
if(((Record*)a)->mou!=((Record *)b)->mou)
return ((Record*)a)->mou-((Record *)b)->mou;
if(((Record*)a)->dd!=((Record *)b)->dd)
return ((Record*)a)->dd-((Record *)b)->dd;
if(((Record*)a)->hh!=((Record *)b)->hh)
return ((Record*)a)->hh-((Record *)b)->hh;
if(((Record*)a)->mm!=((Record *)b)->mm)
return ((Record*)a)->mm-((Record *)b)->mm;
}
else
return tmp;
}
//这个方式很厉害!
void compute_money(Record *r1,Record *r2,int *rate, Res *res){
Record r;
r.dd=r1->dd;
r.hh=r1->hh;
r.mm=r1->mm;
res->money=0;
res->last_time=0;
//get
while(r.dd<r2->dd||r.hh<r2->hh||r.mm<r2->mm){
res->money+=rate[r.hh];
res->last_time++;
r.mm++;
if(r.mm>=60){
r.mm=0;
r.hh++;
if(r.hh>=24){
r.hh=0;
r.dd++;
}
}
}
res->money/=100;
}
int check_ok(Record *r,int index,int n){
char pname[NAME];
int i,flag=0,has=0;
strcpy(pname,r[index].name);
for(i=index;i<n;i++){
if(strcmp(r[i].name,pname)==0){
if(r[i].line[1]=='n'){
flag=1;
}
else if(r[i].line[1]=='f'&&1==flag){
flag=0;
has=1;
}
}
else
break;
}
return has;
}
int main()
{
int rate[HOURS];
int i,n;
for(i=0;i<HOURS;i++)
scanf("%d",&rate[i]);
scanf("%d",&n);
Record *r=new Record
;
for(i=0;i<n;i++){
scanf("%s",r[i].name);
scanf("%d:%d:%d:%d",&r[i].mou,&r[i].dd,&r[i].hh,&r[i].mm);
scanf("%s",r[i].line);
}
qsort(r,n,sizeof(Record),compare_name);
if(n>0){
i=0;
while(i<n){
int has=check_ok(r,i,n);
char pname[NAME];
if(1==has){
printf("%s %02d\n",r[i].name,r[i].mou);
strcpy(pname,r[i].name);
int flag=0;
float t_money=0;
Record rn;
while(strcmp(pname,r[i].name)==0){
if(r[i].line[1]=='n'){
rn.dd=r[i].dd;
rn.hh=r[i].hh;
rn.mm=r[i].mm;
flag=1;
}
else if(r[i].line[1]=='f'&&1==flag){
flag=0;
printf("%02d:%02d:%02d %02d:%02d:%02d",rn.dd,rn.hh,rn.mm,r[i].dd,r[i].hh,r[i].mm);
Res res;
compute_money(&rn,&r[i],rate,&res);
printf(" %d $%.2f\n",res.last_time,res.money);
t_money+=res.money;

}
i++;
}
printf("Total amount: $%.2f\n",t_money);
}
else
i++;
}
}
return 0;
}

参考

兵临城下 1016. Phone Bills (25)-PAT