思路：on off 分两拨 用vector存储 on从大到小 off从小到大
map 将人名和 on 的时间连接起来 人名和off的时间连接起来
然后 两个map开始匹配
最初的想法是off在外面 on在内 若人名不匹配 说明小的那个肯定没有对应的，erase 但是会有运行时错误 所以 反正也匹配不了。。就留着了，
在时间对应的关系上，默认是可以直接匹配的。
但是 只有第一个过了 （15） 其余三个答案错误，暂时没找出来，待解决
#include<iostream>
#include<cstdio>
#include<cmath>
#include<vector>
#include<map>
#include<cstring>
#include<sstream>
#include<algorithm>using namespace std;struct P {int mon;int d;int h;int m;bool flag;P() {flag = false;}friend ostream& operator << (ostream &Out, const P &t) {Out << "mon:" << t.mon << " " << "d:" << t.d<< " " << "h:" << t.h<< " " << "m:" << t.m;return Out;}
};
int mh[25];
bool cmpa(P a, P b) {if(a.mon == b.mon) {if(a.d == b.d) {if(a.h == b.h) {return a.m > b.m;}return a.h > b.h;}return a.d > b.d;}return a.mon > b.mon;
}
bool cmpi(P a, P b) {if(a.mon == b.mon) {if(a.d == b.d) {if(a.h == b.h) {return a.m < b.m;}return a.h < b.h;}return a.d < b.d;}return a.mon < b.mon;
}bool mm[13];
int main() {vector<P> vn;vector<P> vf;map<string, vector<P> > an;map<string, vector<P> > af;mh[24] = 0;for(int i=0; i<24; i++) {cin >> mh[i];mh[24] += mh[i];}int n;cin >> n;P p;string na, j;getchar();for(int i=0; i<n; i++) {stringstream ss;getline(cin, j);ss << j;string tmp;ss >> na;ss >> tmp;ss >> j;sscanf(tmp.c_str(), "%d:%d:%d:%d", &p.mon, &p.d, &p.h, &p.m);//if(j == "on-line")string on = "on-line";if(!strcmp(j.c_str(), on.c_str())) {an[na].push_back(p);} else {af[na].push_back(p);}}for(map<string, vector<P> >::iterator it=an.begin(); it!=an.end(); it++) {
// cout << it->first << endl;vf = it->second;if(vf.size() > 0)sort(it->second.begin(), it->second.end(), cmpa);/*for(int i=0; i<vf.size(); i++) {cout << vf[i] << endl;}*/}//大->小for(map<string, vector<P> >::iterator it=af.begin(); it!=af.end(); it++) {
// cout << it->first << endl;vf = it->second;if(vf.size() > 0) sort(it->second.begin(), it->second.end(), cmpi);/*for(int i=0; i<vf.size(); i++) {cout << vf[i] << endl;}*/}//小->大map<string, vector<P> >::iterator tmp;for(map<string, vector<P> >::iterator it=af.begin(); it!=af.end(); it++) {if(an.size()==0 || af.size() == 0) break;for(map<string, vector<P> >::iterator it1=an.begin(); it1!=an.end();) {if(an.size() == 0) break;if(it->first != it1->first) { //没有对应的if(it->first < it1->first) { //没有对应的it++;continue;} else { /* tmp = it1;an.erase(it1);//因为每次都会在on-line从头开始匹配 erase是为了不在看这个 但是会运行时错误。。it1 = tmp;*/it1++; continue;}} else {// cout << "可以找到对应的的" << endl;for(int i=0; i<=12; i++) {mm[i] = false;}vf = it->second;vn = it1->second;it1++;int total = 0;// for(vector<P>::iterator it3 = vf.begin(); it!)for(int i=0; i<vf.size(); i++) {P tmpf = vf[i];if(tmpf.flag) {continue;}// cout << i << vf[i] << endl;for(int j=0; j<vn.size(); j++) {P tmpo = vn[j];if(tmpo.flag || tmpo.mon != tmpf.mon) {// cout << "mon" << endl;continue;}if(tmpo.d > tmpf.d) {// cout << "d" << endl;continue;} //可以正常计算vf[i].flag = true;vn[j].flag = true;float s = (tmpf.d-tmpo.d)*mh[24]*60;// cout <<mh[24] << "day:" << s << endl;for(int i=tmpo.h; i<tmpf.h; i++) {if(tmpo.m == 0 || i!= tmpo.h)s += mh[i]*60;else s += mh[i]*(60-tmpo.m);}s+= mh[tmpf.h]*tmpf.m;if(s == 0)continue;if(!mm[tmpo.mon]) {mm[tmpo.mon] = true;string tt = it->first;cout << tt;printf(" %02d\n", tmpf.mon);}printf("%02d:%02d:%02d %02d:%02d:%02d %d $%.2f\n", tmpo.d, tmpo.h, tmpo.m, tmpf.d, tmpf.h, tmpf.m, ((tmpf.d-tmpo.d)*24+tmpf.h-tmpo.h)*60+tmpf.m-tmpo.m, (float)s/100);total += s;break;}}if(total)printf("Total amount: $%.2f\n", (float)total/100);}}}return 0;
}/*
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 01:00:07:00 off-line
CYLL 01:01:06:01 off-line
CYJJ 01:01:07:00 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
*/