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  • Codeforces Round #336 (Div. 2) A

    A. Saitama Destroys Hotel
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    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.

    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 ≤ s1 ≤ 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.

    Examples
    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: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

     1 #include<iostream>
     2 #include<cstring>
     3 #include<cstdio>
     4 #include<queue>
     5 #include<stack>
     6 using namespace std;
     7 int n,s;
     8 int a[1005];
     9 int maxn;
    10 int l,r;
    11 int main()
    12 {
    13     scanf("%d %d",&n,&s);
    14     for(int i=0;i<=s;i++)
    15     {
    16         a[i]=0;
    17     }
    18     for(int i=1;i<=n;i++)
    19     {
    20         scanf("%d %d",&l,&r);
    21         if(r>a[l])
    22         a[l]=r;
    23     }
    24     int maxn=-1;
    25     for(int i=s;i>=0;i--)
    26     {
    27         {
    28         if(a[i]>maxn)
    29         maxn=a[i];
    30         else
    31         maxn++;
    32         }
    33     }
    34     cout<<maxn<<endl;
    35     return 0;
    36  } 
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  • 原文地址:https://www.cnblogs.com/hsd-/p/5528931.html
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