hdu 4585　map **

题意：

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.

ｍａｐ的这种用法还没见过呢，下次重拍一遍

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <set>
 #include <map>
 #include <string>
 #include <math.h>
 #include <stdlib.h>
 using namespace std;

 int main()
 {
     map<int,int>mp;
     int n;
     while(scanf("%d",&n) ==  && n)
     {
         mp.clear();
         mp[] = ;
         int u,v;
         while(n--)
         {
             scanf("%d%d",&u,&v);
             printf("%d ",u);
             map<int,int>::iterator it = mp.lower_bound(v);
             if(it == mp.end())
             {
                 it--;
                 printf("%d\n",it->second);
             }
             else
             {
                 int t1 = it->first;
                 int tmp = it->second;
                 if(it != mp.begin())
                 {
                     it--;
                     if(v - it->first <= t1 - v)
                     {
                         printf("%d\n",it->second);
                     }
                     else printf("%d\n",tmp);
                 }
                 else printf("%d\n",it->second);
             }
             mp[v] = u;
         }
     }
     return ;
 }

2015/7/6