标签：include 负数 stat require tis break base another ever
2017-09-24 19:16:38
writer：pprp
题目链接：https://www.jisuanke.com/contest/877
题目如下：
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.
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.
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.
2
1 10 8
20 50 20
3
2 30 5
20 80 20
40 90 40
0
20 60 *
分析：这是一个求解重叠区间最大和的问题，签到题...ai
将起点设为正数，终点设为负数，扫描一遍就可以知道区间中最大值
代码如下：
//ac B
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 105;
int a[maxn];
int main()
{
freopen("in.txt","r",stdin);
int n;
while(cin>>n)
{
if(n==0)
{
cout<<"*"<<endl;
break;
}
memset(a,0,sizeof(a));
while(n--)
{
int s,t,k;
cin>>s>>t>>k;
a[s]+=k;
a[t]-=k;
}
int ans=0,temp=0;
for(int i=1; i<=100; i++)
{
temp+=a[i];
ans=max(ans,temp);
}
cout<<ans<<endl;
}
return 0;
}
标签：include 负数 stat require tis break base another ever
原文地址：http://www.cnblogs.com/pprp/p/7588171.html
