1016. Phone Bills 解析

直接把时间化成以秒为单位 排序 匹配 计算就好。

#include <iostream>#include <string>#include <vector>#include <set>#include <algorithm>using namespace std;struct Node {string name;string time;string state;};int char2int(char i) {int result = 0;switch (i){case '1': result = 1; break;case '2': result = 2; break;case '3': result = 3; break;case '4': result = 4; break;case '5': result = 5; break;case '6': result = 6; break;case '7': result = 7; break;case '8': result = 8; break;case '9': result = 9; break;case '0': result = 0; break;default:break;}return result;}int str2day(string s) {int day = 0;day = char2int(s[3]) * 10 + char2int(s[4]);return day;}int str2hour(string s) {int hour = 0;hour = char2int(s[6]) * 10 + char2int(s[7]);return hour;}int str2minute(string s) {int minute = 0;minute = char2int(s[9]) * 10 + char2int(s[10]);return minute;}int CalRate(string time1, string time2, vector <int> rate) {//M1 < M2int D1, D2, H1, H2, M1, M2;D1 = str2day(time1);D2 = str2day(time2);H1 = str2hour(time1);H2 = str2hour(time2);M1 = str2minute(time1);M2 = str2minute(time2);int Sum = 0;if (D1 == D2) { //同一天if (H1 == H2) {//同一小时Sum += (M2 - M1)* rate[H1];return Sum;}else{//不同小时Sum += (60 - M1) * rate[H1];Sum += M2 * rate[H2];for (int i = H1+1; i < H2; i++)Sum += 60 * rate[i];return Sum;}}else{ //不同天Sum += (60 - M1)* rate[H1];for (int i = H1 + 1; i < 24; i++) {Sum += 60 * rate[i];}Sum += M2 * rate[H2];for (int i = 0; i < H2; i++) {Sum += 60 * rate[i];}for (int i = D1 + 1; i < D2; i++) {for (int j = 0; j < 24; j++) {Sum += rate[j] * 60;}}return Sum;}}void PrintTime(string s) {for (int i = 3; i <= 10; i++) {cout << s[i];}}void TotalRate(vector <Node> list ,vector <int> rate) { //对一个用户的账单进行生成bool tag = false;//是否匹配bool Head = true;//又没有显示Headbool HaveBill = false;float SumRate = 0;int time = 0;Node pre; //匹配前//cout << list[0].name << " " << list[0].time[0] << list[0].time[1] << endl;for (int i = 0; i < list.size(); i++) {if (!tag && list[i].state == "off-line")//前面没有on 却有off 忽略continue;else if (!tag && list[i].state == "on-line") {//前面没有on 后面weion pre等于该数tag = true;pre = list[i];}else if (tag && list[i].state == "on-line") { //前面有on 后面还有on 更新prepre = list[i];}else {//匹配成功if(Head){cout << list[i].name << " " << list[i].time[0] << list[i].time[1] << endl;Head = false;HaveBill = true;}tag = false;int PreTime = str2day(pre.time) * 24 * 60 + str2hour(pre.time) * 60 + str2minute(pre.time);int Time = str2day(list[i].time) * 24 * 60 + str2hour(list[i].time) * 60 + str2minute(list[i].time);int gapTime = Time - PreTime;float tempRate = CalRate(pre.time, list[i].time, rate);float tR = tempRate / 100;SumRate += tR;PrintTime(pre.time); cout << " ";PrintTime(list[i].time);cout << " " << gapTime << " $";printf("%.02f", tR);cout << endl;}}if (HaveBill) {cout << "Total amount: $";printf("%.2f", SumRate);cout << endl;}}bool cmp(Node N1 ,Node N2) {if (N1.name < N2.name)return true;else if (N1.name == N2.name && str2day(N1.time) < str2day(N2.time))return true;else if (N1.name == N2.name && str2day(N1.time) == str2day(N2.time) && str2hour(N1.time) < str2hour(N2.time))return true;else if (N1.name == N2.name && str2day(N1.time) == str2day(N2.time) && str2hour(N1.time) == str2hour(N2.time) && str2minute(N1.time) < str2minute(N2.time))return true;elsereturn false;}int main() {vector <int> rate;int tempRate;for (int i = 0; i < 24; i++) {cin >> tempRate;rate.push_back(tempRate);}int N;//记录数cin >> N;Node * record = new Node[N];set<string> r;for (int i = 0; i < N; i++) {cin >> record[i].name >> record[i].time >> record[i].state;r.insert(record[i].name);}sort(record, record + N, cmp);//对记录进行排序vector <Node> * List = new vector<Node>[r.size()]; //对不同用户进行分组string temp = record[0].name;int tempi = 0;for (int i = 0; i < N; i++) {//分组if (temp == record[i].name)List[tempi].push_back(record[i]);else{tempi++;temp = record[i].name;List[tempi].push_back(record[i]);}}//for (int i = 0; i < r.size(); i++) {//for (int j = 0; j < List[i].size(); j++)//cout << List[i][j].name << " " << List[i][j].time << " " << List[i][j].state << endl;//}for (int i = 0; i < r.size(); i++) {TotalRate(List[i], rate);}return 0;}