***A. Lala Land and Apple Trees***
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 xi and has ai 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 xi, ai ( - 105 ≤ xi ≤ 105, xi ≠ 0, 1 ≤ ai ≤ 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题目大意：就是以0为原点，从左往右折返跑，直到有一边没有数，
看最大的数是多少
样例解释；
input
3
-2 2
1 4
-1 3
output
9
ret == 0；
先往左跑 ret == 0+3 == 3;
再往右跑 ret == 3+4 == 7;
再往左跑 ret == 7+2 == 9;
所以 ｒｅｔ ＝＝ ９；
解体思路：
排一下序就行，然后找一下正数多还是负数多，再根据题意判断一下就ｏｋ了
上代码：
/*
Date : 2015-09-03 下午
Author : ITAKING
Motto :
今日的我要超越昨日的我，明日的我要胜过今日的我；
以创作出更好的代码为目标，不断地超越自己。
*/
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
pair <int, int> arr[105];
int main()
{
int m, ans=0;
scanf("%d", &m);
for(int i=0; i<m; i++)
{
cin>>arr[i].first>>arr[i].second;
if(arr[i].first < 0)
ans++;
}
sort(arr, arr+m);
int ret = 0;
if(2*ans < m)
for(int i=0; i<min(2*ans+1, m); i++)
ret += arr[i].second;
else
for(int i=max(0,2*ans-m-1); i<m; i++)
ret += arr[i].second;
cout<<ret<<endl;
return 0;
}
