Problem Description
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
3
2 1
3 3
4 2
0
Sample Output
2 1
3 2
4 2
Source
题意：讲的是一群无聊的和尚在比武，当来了一个新和尚时要挑选一个战斗力与他最接近的老和尚比武，
如果有两个老和尚和尚与他的战斗力的绝对值相等，则选择战斗力比他小的。
输出这个新和尚的编号和与他比武的老和尚的编号
直接用map标记就行了，
不过还是WA了好几次，说明自己还是很差啊！
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<vector>
5 #include<set>
6 #include<algorithm>
7 #include<cmath>
8 #include<stdlib.h>
9 #include<map>
10 using namespace std;
11 #define N 100006
12 int n;
13 map<int,int>mp;
14 map<int,int>::iterator it,it2;
15 int main()
16 {
17 while(scanf("%d",&n)==1 && n)
18 {
19 mp.clear();
20 mp[1000000000]=1;
21 for(int i=0;i<n;i++)
22 {
23 int x,y;
24 scanf("%d%d",&x,&y);
25 it=mp.lower_bound(y);
26 if(it==mp.begin())//当新的只能排在第一位时，直接输出第一个
27 {
28 printf("%d %d\n",x,it->second);
29 }
30 else
31 {
32 it2=it;
33 it--;
34 if(abs(y-it->first)>abs(y-it2->first))
35 printf("%d %d\n",x,it2->second);
36 else
37 printf("%d %d\n",x,it->second);
38 }
39 mp[y]=x;
40 }
41 }
42 return 0;
43 }