PAT (Advanced Level) 1016 Phone Bills (map,set,pair)

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这种比较麻烦的数据处理考察语法吧感觉。

map，set这两个结构会自动进行排序。因此对此题很便利。

总的来说这题如果语法，库都比较熟悉就很简单，由于有些结构没怎么用过，因此照着网上的代码敲了遍。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<map>
#include<set>
using namespace std;
int rate[25];
int n;
struct Detail{
	int dd, hh, mm;
	string status;
	Detail(){}
	Detail(int dd, int hh, int mm, string status) : dd(dd), hh(hh),mm(mm), status(status){}
	bool operator < (const Detail &b) const {
		if(dd != b.dd)
			return dd < b.dd;
		else{
			if(hh != b.hh)
				return hh < b.hh;
			else
				return mm < b.mm;
		}
	}
};
map<pair<string,int>,set<Detail> > mp;

int calculate(const Detail &a, const Detail &b){
	printf("%02d:%02d:%02d %02d:%02d:%02d ", a.dd, a.hh, a.mm, b.dd, b.hh, b.mm);
	int cost = -a.mm*rate[a.hh] + b.mm*rate[b.hh];//把前面的时间归零，后面的时间加上
	int time = -a.mm + b.mm;//时间同理 
	int tmph = a.hh, tmpd = a.dd;
	//先处理天数，前面归零的时间，也在这里被补上了 
	while(tmpd < b.dd){
		do{
			cost += 60*rate[tmph];
			tmph++;
			tmph %= 24;
			time += 60;
		}while(tmph != 0);
		tmpd++;
	}
	//同一天了，此时 tmph = 0 
	while(tmph < b.hh){
		cost += 60*rate[tmph];
		time += 60; 
		tmph++;
	} 
	
	printf("%d $%.2f\n",time,(float)cost/100);
	return cost;
}
void print(string name, int month, set<Detail> &st){
	int total = 0, online = 0;
	Detail tmp;
	for(set<Detail>::iterator it= st.begin(); it != st.end(); it++){
		if(it->status == "on-line"){
			tmp = *it;
			if(!online){
				online = 1;
			}
		}
		else{
			if(online){
				if(total == 0){
					cout << name;
					printf(" %02d\n",month);
				}
				total += calculate(tmp, *it);
				online = 0;
			}
		}
	}
	if(total)
		printf("Total amount: $%.2f\n",(float)total/100);
}
int main(){
	for(int i = 0; i < 24; i++){
		scanf("%d",&rate[i]);
	}
	scanf("%d",&n);
	for(int i = 0; i < n; i++){
		string name, status;
		int M,d,h,m;
		cin >> name;
		scanf("%d:%d:%d:%d",&M,&d,&h,&m);
		cin >> status;
		mp[make_pair(name,M)].insert(Detail(d,h,m,status));
	}
	for(map<pair<string,int>,set<Detail> >::iterator it = mp.begin(); it != mp.end(); it++){
		print(it->first.first, it->first.second, it->second);	
	}
	
	return 0;
}