2017 ACM-ICPC 亚洲区（南宁赛区）网络赛-- B. Train Seats Reservation-(区间更新，单点查询思想)

You are given a list of train stations, say from the station 11 to
the station 100100.
The passengers can order several tickets from one station to another before the train leaves the station one. We will issue one train from the station 11 to
the station 100100 after
all reservations have been made. Write a program to determine the minimum number of seats required for all passengers so that all reservations are satisfied without any conflict.

Note that one single seat can be used by several passengers as long as there are no conflicts between them. For example, a passenger from station 11 to
station 1010 can
share a seat with another passenger from station 3030to 6060.

Input Format

Several sets of ticket reservations. The inputs are a list of integers. Within each set, the first integer (in a single line) represents the number of orders, nn,
which can be as large as 10001000.
After nn,
there will be nn lines
representing the nn reservations;
each line contains three integers s,
t, ks,t,k,
which means that the reservation needs kk seats
from the station ss to
the station tt .These
ticket reservations occur repetitively in the input as the pattern described above. An integer n
= 0n=0 (zero)
signifies the end of input.

Output Format

For each set of ticket reservations appeared in the input, calculate the minimum number of seats required so that all reservations are satisfied without conflicts. Output a single star
'*' to signify the end of outputs.

样例输入

2
1 10 8
20 50 20
3
2 30 5
20 80 20
40 90 40
0

样例输出

20
60
*

此题目想到了树状数组，但是只要模拟区间更新，单点查询的思想就行了

代码：

#include<iostream>
#include<string>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<iomanip>
#include<queue>
#include<cstring>
#include<map>
#include<vector>
using namespace std;
typedef long long ll;
int a[105],n;
int main()
{
int s,t,k;
while(scanf("%d",&n)!=EOF)
{
if(n==0) {printf("*"); break;}
memset(a,0,sizeof(a));
while(n--)
{
scanf("%d%d%d",&s,&t,&k);
a[s]+=k;
a[t]-=k;
}
int maxn=0,sum=0;
for(int i=1;i<=100;i++)
{
sum+=a[i];
if(sum>maxn) maxn=sum;
}
printf("%d\n",maxn);
}
return 0;
}