Saitama Destroys Hotel

最新推荐文章于 2020-05-07 17:42:10 发布

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最新推荐文章于 2020-05-07 17:42:10 发布

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分类专栏：水题文章标签：水题模拟

本文链接：https://blog.youkuaiyun.com/qq_30076791/article/details/50442425

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本文讨论了Saitama不慎破坏酒店后，Genos为赔偿而操作电梯的问题。文章详细阐述了一个特殊的电梯系统，该系统只能向下移动，且拥有无限容量。通过输入参数包括乘客数量、顶部楼层数以及每个乘客的到达时间与楼层，算法旨在计算将所有乘客带到地面所需的时间。通过实例分析，展示了如何通过最优调度策略减少等待时间。

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C - Saitama Destroys Hotel

Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status

Description

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 tos and 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.

Input

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 f_i and t_i (1 ≤ f_i ≤ s, 1 ≤ t_i ≤ 1000) — the floor and the time of arrival in seconds for the passenger number i.

Output

Print a single integer — the minimum amount of time in seconds needed to bring all the passengers to floor 0.

Sample Input

Input

Output

Input

Output

Hint

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.

#include<stdio.h>
int main()
{
    int n,s;
    scanf("%d%d",&n,&s);
    int f[1005],t[1005];
    for(int i=0; i<n; i++)
        scanf("%d%d",&f[i],&t[i]);
    int sum=s-f[n-1];
    for(int i=n-1; i>=1; i--)
    {
        if(sum<t[i])
        {
            sum+=t[i]-sum;//等待
        }
        sum+=f[i]-f[i-1];//到达下一层楼
    }
    if(sum<t[0])
        sum+=t[0]-sum;
    sum+=f[0];
    printf("%d\n",sum);
    return 0;
}