1 second
256 megabytes
standard input
standard output
Saitama accidentally destroyed a hotel again. To repay the hotel company, Genos has volunteered to operate an elevator in one of its other hotels. The elevator is special — it starts on the top floor, can only move down, and has infinite capacity. Floors are numbered from 0 to sand elevator initially starts on floor s at time 0.
The elevator takes exactly 1 second to move down exactly 1 floor and negligible time to pick up passengers. Genos is given a list detailing when and on which floor passengers arrive. Please determine how long in seconds it will take Genos to bring all passengers to floor 0.
The first line of input contains two integers n and s (1 ≤ n ≤ 100, 1 ≤ s ≤ 1000) — the number of passengers and the number of the top floor respectively.
The next n lines each contain two space-separated integers fi and ti (1 ≤ fi ≤ s, 1 ≤ ti ≤ 1000) — the floor and the time of arrival in seconds for the passenger number i.
Print a single integer — the minimum amount of time in seconds needed to bring all the passengers to floor 0.
3 7
2 1
3 8
5 2
11
5 10
2 77
3 33
8 21
9 12
10 64
79
In the first sample, it takes at least 11 seconds to bring all passengers to floor 0. Here is how this could be done:
1. Move to floor 5: takes 2 seconds.
2. Pick up passenger 3.
3. Move to floor 3: takes 2 seconds.
4. Wait for passenger 2 to arrive: takes 4 seconds.
5. Pick up passenger 2.
6. Go to floor 2: takes 1 second.
7. Pick up passenger 1.
8. Go to floor 0: takes 2 seconds.
This gives a total of 2 + 2 + 4 + 1 + 2 = 11 seconds.
题意：s层楼 电梯1秒下一层 并且只能向下 n个人 在不同楼层 以及到达该楼层的时间
题解；到达的0层的时间
水
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<stack>
using namespace std;
int n,s;
int a[];
int maxn;
int l,r;
int main()
{
scanf("%d %d",&n,&s);
for(int i=;i<=s;i++)
{
a[i]=;
}
for(int i=;i<=n;i++)
{
scanf("%d %d",&l,&r);
if(r>a[l])
a[l]=r;
}
int maxn=-;
for(int i=s;i>=;i--)
{
{
if(a[i]>maxn)
maxn=a[i];
else
maxn++;
}
}
cout<<maxn<<endl;
return ;
}