Shaolin

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 3205    Accepted Submission(s): 1360

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

32 13 34 20

 

Sample Output

2 13 24 2

 

Source

2013ACM-ICPC杭州赛区全国邀请赛

题意大意：很多人想进少林寺，少林寺最开始只有一个和尚，每个人有有一个武力值，若这个人想进少林，必须和比他先进去的人比武并且武力值最接近他的比武，如果有相同的则选择武力值比他小的，问当他进去的时候要和哪个和尚比武

lower_bound（）总结：

当参数 key 没有在容器 key的范围内：
1. 小于容器key uper_bound， lower_bound 都将返回 begin.
2. 大于容器key uper_bound, lower_bound 都将返回 end（注意end已经超界了）
当参数key 在容器key 范围内：
1. 参数 key == 容器key. lower_bound 将返回当前key 的iterator， uper_bound 将返回下一个元素的iterator.
2. 参数 key 不等于容器key，且在范围内， lower_bound将返回比参数key 大的且相邻的容器key的iterator
3 如果 Key等于 begin 或等于 end，将返回begin 或end

#include <iostream>  #include <algorithm>  #include <cstdio>  #include <map>  #include <algorithm>  using namespace std;    int main()  {        int n;      while(scanf("%d",&n)!= EOF && n)      {          map<int,int> mp;          mp[1e9] = 1;            while(n--)          {              int k,g;              scanf("%d%d",&k,&g);              map<int,int>::iterator i = mp.lower_bound(g);              if(i == mp.begin())  printf("%d %d\n",k,i->second);              else              {                  map<int,int>::iterator a = i, b= --i;                  if((a->first - g) >= (g -b->first) )                  {                      printf("%d %d\n",k,b->second);                  }                  else printf("%d %d\n",k,a->second);              }              mp[g] = k;            }      }      return 0;  }