Train Seats Reservation

You are given a list of train stations, say from the station  to the station 100.

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 1 to the station 100 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 1 to station 10 can share a seat with another passenger from station 30 to 60.

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, nnn, which can be as large as 1000. After nnn, there will be nnn lines representing the nnn reservations; each line contains three integers s,t,k, which means that the reservation needs k seats from the station s to the station t .These ticket reservations occur repetitively in the input as the pattern described above. An integer n=0n = 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
*

题目来源

2017 ACM-ICPC 亚洲区（南宁赛区）网络赛

由于最多只有100个站，所以可以暴力。

#include <iostream>
#include <stdio.h>
#include <string.h>
#define ll long long
using namespace std;
const int N = 110;
int a[N];
int main() {
    int n;
    while(scanf("%d", &n) &&n) {
        int s, t, k;
        for(int i = 1; i <= n; i ++) {
            scanf("%d %d %d", &s, &t, &k);
            for(int j = s; j < t; j ++) {
                a[j] += k;
            }
        }
        int ans = 0;
        for(int i = 1; i <= 100; i ++) {
            ans = max(ans, a[i]);
        }
        printf("%d\n",ans);
        memset(a, 0, sizeof(a));
    }
    printf("*\n");
    return 0;
}