CF 558A (Lala Land and Apple Trees-Brute Force)
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 exactlyNApple trees. Tree numberIIs located in a positionXIAnd hasAIApples growing on it. Amr wants to collect apples from the apple trees. Amr currently stands inXCursor = cursor 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 numberN(1 digit ≤ DigitNLimit ≤ limit 100), the number of apple trees in Lala Land.
The followingNLines contains two integers eachXI,AI(Cost-limit 105 limit ≤ limitXILimit ≤ limit 105,XI =0 0, 1 ≤AILimit ≤ limit 105), representing the position ofI-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 51 5
Output
10
Input
3-2 21 4-1 3
Output
9
Input
31 93 57 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 leftXDirection = direction-direction 1, collect apples from there, then the direction will be reversed, Amr has to goXRows = Limit 1, collect apples from there, then the direction will be reversed and Amr goes to the final treeXBytes = bytes-cached 2.
In the third sample test the optimal solution is to go rightXRows = Limit 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.
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using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i
=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p]) #define Lson (x<<1)#define Rson ((x<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (100000007)#define MAXN (1000)typedef long long ll;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return (a-b+(a-b)/F*F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int n;pair
p[MAXN];int main(){//freopen(A.in,r,stdin);//freopen(.out,w,stdout);cin>>n;For(i,n) scanf(%d%d,&p[i].first,&p[i].second);sort(p+1,p+1+n);p[0]=p[n+1]=make_pair(-INF,-INF);int p2=1;while(p2<=n&&p[p2].first<0) ++p2;if (p2>n){cout<
1) ans+=p[l-1].second;if (l>1&&r