codeforces 558A. Lala Land and Apple Trees

本是古典何须时尚 2022-08-02 07:28 145阅读 0赞

>                                                       A. Lala Land and Apple Trees
> 
>                                                                  time limit per test
> 
>                                                                         1 second
> 
>                                                          memory limit per test
> 
>                                                               256 megabytes
> 
>                                                                    input
> 
>                                                               standard input
> 
>                                                                      output
> 
>                                                              standard output
> 
> Amr lives in Lala Land. Lala Land is a very beautiful country that is located on a coordinate line. Lala Land is famous with its apple trees growing everywhere.
> 
> Lala Land has exactly *n* apple trees. Tree number *i* is located in a position *x**i* and has *a**i* apples growing on it. Amr wants to collect apples from the apple trees. Amr currently stands in *x* = 0 position. At the beginning, he can choose whether to go right or left. He'll continue in his direction until he meets an apple tree he didn't visit before. He'll take all of its apples and then reverse his direction, continue walking in this direction until he meets another apple tree he didn't visit before and so on. In the other words, Amr reverses his direction when visiting each new apple tree. Amr will stop collecting apples when there are no more trees he didn't visit in the direction he is facing.
> 
> What is the maximum number of apples he can collect?
> 
> Input
> 
> The first line contains one number *n* (1 ≤ *n* ≤ 100), the number of apple trees in Lala Land.
> 
> The following *n* lines contains two integers each *x**i*, *a**i* ( - 105 ≤ *x**i* ≤ 105, *x**i* ≠ 0, 1 ≤ *a**i* ≤ 105), representing the position of the *i*\-th tree and number of apples on it.
> 
> It's guaranteed that there is at most one apple tree at each coordinate. It's guaranteed that no tree grows in point 0.
> 
> Output
> 
> Output the maximum number of apples Amr can collect.
> 
> Sample test(s)
> 
> Input
> 
>     2
>     -1 5
>     1 5
> 
> Output
> 
>     10
> 
> Input
> 
>     3
>     -2 2
>     1 4
>     -1 3
> 
> Output
> 
>     9
> 
> Input
> 
>     3
>     1 9
>     3 5
>     7 10
> 
> Output
> 
>     9
> 
> Note
> 
> In the first sample test it doesn't matter if Amr chose at first to go left or right. In both cases he'll get all the apples.
> 
> In the second sample test the optimal solution is to go left to *x* =  - 1, collect apples from there, then the direction will be reversed, Amr has to go to *x* = 1, collect apples from there, then the direction will be reversed and Amr goes to the final tree *x* =  - 2.
> 
> In the third sample test the optimal solution is to go right to *x* = 1, collect apples from there, then the direction will be reversed and Amr will not be able to collect anymore apples because there are no apple trees to his left.
> 
>   
> 
> 
> 把正数跟负数存放在不同的数组中分别进行排序，如果正数个数大于负数个数，肯定先往正数一边走最优，如果负数个数大于正数个数，肯定先往负数一边走最优，如果个数相等，那么苹果全部都能吃到，把数组中的元素全加起来就行。  
> 
> 
>   
> 
> 
>     #include<map>
>     #include<vector>
>     #include<cstdio>
>     #include<iostream>
>     #include<cstring>
>     #include<string>
>     #include<algorithm>
>     #include<cmath>
>     #include<stack>
>     #include<queue>
>     #include<set>
>     #define inf 0x3f3f3f3f
>     #define mem(a,x) memset(a,x,sizeof(a))
>     
>     using namespace std;
>     
>     typedef long long ll;
>     typedef pair<int,int> pii;
>     
>     inline int in()
>     {
>         int res=0;char c;
>         while((c=getchar())<'0' || c>'9');
>         while(c>='0' && c<='9')res=res*10+c-'0',c=getchar();
>         return res;
>     }
>     struct st
>     {
>         int pos,x;
>     }a[111],b[111];
>     
>     bool cmp(st a, st b)
>     {
>         return a.pos<b.pos;
>     }
>     int main()
>     {
>         int n=in();
>     
>         int c1=0,c2=0;
>         for(int i=0;i<n;i++)
>         {
>             int t1,t2;
>             scanf("%d%d",&t1,&t2);
>             if(t1>0) a[c1].pos=t1,a[c1++].x=t2;
>             else  b[c2].pos=t1,b[c2++].x=t2;
>     
>         }
>         sort(a,a+c1,cmp);
>         sort(b,b+c2,cmp);
>         int ans=0;
>     
>         if(c2>c1)
>         {
>             for(int i=0;i<c1;i++)
>             {
>                 ans += a[i].x;
>             }
>             for(int i=c2-1;i>=c2-c1-1;i--)
>             {
>                 ans += b[i].x;
>             }
>         }
>         else if(c1>c2)
>         {
>             for(int i=0;i<c2;i++)
>             {
>                 ans += b[i].x;
>             }
>             for(int i=0;i<=c2;i++)
>             {
>                 ans += a[i].x;
>             }
>         }
>         else
>         {
>             for(int i=0;i<c1;i++)
>             {
>                 ans+=a[i].x+b[i].x;
>             }
>         }
>         cout<<ans<<endl;
>     
>         return 0;
>     }
> 
>   
>   
> 
> 
>   
>