Codeforces Round #336 (Div. 2) A. Saitama Destroys Hotel 模拟

A. Saitama Destroys Hotel

 

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 s 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 fi and ti (1 ≤ fi ≤ s, 1 ≤ ti ≤ 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 test(s)

input

3 7
2 1
3 8
5 2

output

11

input

5 10
2 77
3 33
8 21
9 12
10 64

output

79

Note

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：电梯最开始在最顶楼，里面载了0个人，给出下面n个人会在第几楼第几秒出现，让你计算 电梯从其实位置接到n个人 并且回到第0层楼 所需的最少时间

题解：  因为你必然经过每一层楼，我们找到接哪个人所需时间最多就好了，遍历一遍

//meek///#include<bits/stdc++.h>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include<iostream>
#include<bitset>
using namespace std ;
#define mem(a) memset(a,0,sizeof(a))
#define pb push_back
#define fi first
#define se second
#define MP make_pair
typedef long long ll;
 
const int N = ;
const int inf = 0x3f3f3f3f;
const int MOD = ;
const double eps = 0.000001;
 
int n,s;
struct ss {
 int f,t,n,s;
}a[N];
int cmp (ss s1,ss s2) {
  return s1.f>s2.f;
}
int main() {
    scanf("%d%d",&n,&s);
    int an=-;
    for(int i=;i<=n;i++) {
        scanf("%d%d",&a[i].f,&a[i].t);
            an=max(an,a[i].f+max(a[i].t,s-a[i].f));
    }
    cout<<an<<endl;
    return ;
}

代码