题意：

少林寺有n+1个和尚，他们都有一个独有的编号和战斗力值，当一个年轻人通过所有考试并被宣布为少林的新僧人时，将会有一场战斗，作为欢迎的一部分。新和尚必须与一位战斗等级最接近他的战斗等级的老和尚战斗。如果有两个老僧人满足这个条件，新僧侣将采取战斗等级低于他的僧侣与他对打。现在保证输入是按照编号顺序升序输入的，要求按顺序输出每一组战斗的双方编号，先输出新和尚的后输出老和尚的。（第一个和尚编号是1，战斗力是1000000000）

题目：

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

分析：

这个题充分体现出了map的优势，因为map存储的时候是按照关键字升序存储的，所以如果按照战斗力值作为关键字建立map的话，找到此战力值的前驱和后继就是可能要与新和尚对打的老和尚了
map的专项知识点看我的另一篇博客：https://blog.csdn.net/zeng_jun_yv/article/details/104704897

AC代码：

#include<bits/stdc++.h>
using namespace std;
map<int,int>a;
int n,x,y;
int main()
{while(~scanf("%d",&n)&&n){a.clear();a[1000000000]=1;for(int i=1; i<=n; i++){cin>>x>>y;a[y]=x;map<int,int>::iterator it=a.find(y);if(it==a.begin())printf("%d %d\n",x,(++it)->second);else if(it==a.end())printf("%d %d\n",x,(--it)->second);else if((abs((++it)->first)-(--it)->first)<abs((--it)->first-(++it)->first))printf("%d %d\n",x,(++it)->second);elseprintf("%d %d\n",x,(--it)->second);}}
}