HDU 4585 Shaolin (STL)

题意：
给出从N-1个点的id（2~N）和权值并保证每个点的id个权值不一样，每输入一个点，输出先前给出的点中权值最接近的点的id，若有多个点满足条件输出id较小的那个。

起初的第一个点权值为 1e9。

思路：
题意简单来说无非就是做一个在线数据的插入和查询，数据的数量为1e5次，数据的最大值为5e6。

根据题意很容易想到时AVL树，stl实现即可。也可以用离散化线段树完成，但十分复杂，在这就不给出了。

代码：

（STL）

#include <bits/stdc++.h>using namespace std;
const int MAXN=1000000000;
typedef struct Node{int x;int y;Node(int xx=0,int yy=0){x=xx;y=yy;}bool operator < (const Node &a)const{if(a.y==y)return a.x<x;return a.y>y;}
}Node;
int main()
{set <Node> ans;int n,fans;while(~scanf("%d",&n)&&n){ans.clear();ans.insert(Node(1,MAXN));Node t;scanf("%d%d",&t.x,&t.y);cout<<t.x<<' '<<1<<endl;ans.insert(t);for(int i=1;i<n;i++){scanf("%d%d",&t.x,&t.y);set <Node>::iterator it=ans.upper_bound(t);if(it==(ans.begin()++)){//cout<<it->x<<' '<<it->y<<endl;fans=it->x;}else{Node ans1=*it;Node ans2=*(--it);//cout<<"ans1 :"<<ans1.x<<' '<<ans1.y<<endl;//cout<<"ans2 :"<<ans2.x<<' '<<ans2.y<<endl;if(abs(ans1.y-t.y)<abs(ans2.y-t.y)){fans=ans1.x;}else{fans=ans2.x;}}cout<<t.x<<' '<<fans<<endl;ans.insert(t);}}
}

Shaolin temple is very famous for its Kongfu monks.A lot of young men go to Shaolin temple every year, trying to be a monk there. The master of Shaolin evaluates a young man mainly by his talent on understanding the Buddism scripture, but fighting skill is also taken into account.
When a young man passes all the tests and is declared a new monk of Shaolin, there will be a fight , as a part of the welcome party. Every monk has an unique id and a unique fighting grade, which are all integers. The new monk must fight with a old monk whose fighting grade is closest to his fighting grade. If there are two old monks satisfying that condition, the new monk will take the one whose fighting grade is less than his.
The master is the first monk in Shaolin, his id is 1，and his fighting grade is 1,000,000,000.He just lost the fighting records. But he still remembers who joined Shaolin earlier, who joined later. Please recover the fighting records for him.

Input

There are several test cases.
In each test case:
The first line is a integer n (0 <n <=100,000),meaning the number of monks who joined Shaolin after the master did.(The master is not included).Then n lines follow. Each line has two integer k and g, meaning a monk's id and his fighting grade.( 0<= k ,g<=5,000,000)
The monks are listed by ascending order of jointing time.In other words, monks who joined Shaolin earlier come first.
The input ends with n = 0.

Output

A fight can be described as two ids of the monks who make that fight. For each test case, output all fights by the ascending order of happening time. Each fight in a line. For each fight, print the new monk's id first ,then the old monk's id.

Sample Input

Sample Output

2 1
3 2
4 2

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