引入地址让函数内可以改变全局变量。
利用next来标记具体人物
然后on在next 内移动，寻找配对的on和off输出

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;


	





typedef struct node
{
	char name[100];
	int MM;
	int dd;
	int hh;
	int mm;
	bool status;
}record;


	int toll[24];
	record R[10000];


bool cmp(record a,record b)
{
	int s = strcmp(a.name,b.name);
	if(s!=0)
		return s<0;
	else if(a.MM !=b.MM)
		return a.MM<b.MM;
	else if(a.dd != b.dd)
		return a.dd<b.dd;
	else if(a.hh != b.hh)
		return a.hh<b.hh;
	else if(a.mm != b.mm)
		return a.mm<b.mm;	
} 
void getans(int on,int off,int &time,int &money)
{
	record temp = R[on];
	while(temp.dd < R[off].dd || temp.hh < R[off].hh || temp.mm < R[off].mm)
	{
		time ++;
		money += toll[temp.hh];
		temp.mm ++;
		if(temp.mm >=60)
		{
			temp.mm = 0;
			temp.hh++;
		}
		if(temp.hh >=24)
		{
			temp.hh =0;
			temp.dd++;
		}
	}
}



int main(int argc, char** argv) {
	

	int n=0;
	char Sta[1000];
	int i =0,flag = 0;	
	int P[100] = {0};
	int index =0;	
	int j =0;	
	
	for(int i =0;i<24;i++)
	{
		scanf("%d",&toll[i]);
	}
	

	
	scanf("%d",&n);
	for(i = 0;i<n;i++)
	{
		scanf("%s",R[i].name);

		scanf("%d:%d:%d:%d",&R[i].MM,&R[i].dd,&R[i].hh,&R[i].mm);

		scanf("%s",Sta);
		if(strcmp(Sta,"on-line") == 0)
		{
			R[i].status = true;
		}
		else
			R[i].status = false;
	}
	
	sort(R,R+n,cmp);
	
	
	
	int on = 0,next,off;

	while(on<n)
	{
		int needp = 0;
		next = on;

			while(next <n &&strcmp(R[on].name,R[next].name) == 0)
			{
				if(R[next].status == true && needp ==0)
					needp = 1;
				else if(R[next].status == false && needp ==1)
					needp = 2;
				next ++;
			}
			if(needp != 2)
			{
				on = next;
				continue;
			}
			
			int All = 0;
			printf("%s %02d\n",R[on].name,R[on].MM);
			
			while(on<next)
			{
				while(on<next - 1 &&
				!(R[on].status == true && R[on +1].status == false))
				{
					on ++;
				}
				off = on +1;
				if(off == next)
				{
					on = next;
					break;
					
				}
				printf("%02d:%02d:%02d ",R[on].dd,R[on].hh,R[on].mm);
				printf("%02d:%02d:%02d ",R[off].dd,R[off].hh,R[off].mm);
				int time=0,money =0;
				getans(on,off,time,money);
				printf("%d $%.2f\n",time,money/100.0);
				
				All +=money;
				
				on = off +1;
			}
			printf("Total amount: $%.2f\n",All/100.0) ;
		
	}	



	
	return 0;
}
在这里插入代码片

题目以后还得再做一遍，自己水平太低，看了好几遍算法笔记才写出来。