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

 1 #include <stdio.h>

 2 #include <string.h>

 3 #include <iostream>

 4 #include <algorithm>

 5 #include <vector>

 6 #include <queue>

 7 #include <set>

 8 #include <map>

 9 #include <string>

10 #include <math.h>

11 #include <stdlib.h>

12 using namespace std;

13 

14 int main()

15 {

16     map<int,int>mp;

17     int n;

18     while(scanf("%d",&n) == 1 && n)

19     {

20         mp.clear();

21         mp[1000000000] = 1;

22         int u,v;

23         while(n--)

24         {

25             scanf("%d%d",&u,&v);

26             printf("%d ",u);

27             map<int,int>::iterator it = mp.lower_bound(v);

28             if(it == mp.end())

29             {

30                 it--;

31                 printf("%d\n",it->second);

32             }

33             else

34             {

35                 int t1 = it->first;

36                 int tmp = it->second;

37                 if(it != mp.begin())

38                 {

39                     it--;

40                     if(v - it->first <= t1 - v)

41                     {

42                         printf("%d\n",it->second);

43                     }

44                     else printf("%d\n",tmp);

45                 }

46                 else printf("%d\n",it->second);

47             }

48             mp[v] = u;

49         }

50     }

51     return 0;

52 }

2015/7/6