A--HDU4585，hdu4585
Shaolin
Time Limit: 1000 MS Memory Limit: 32768 KB
64-bit integer IO format: %I64d , %I64u Java class name: Main
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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
2013ACM-ICPC杭州賽區全國邀請賽
題目大意：
有n個人要成為少林寺弟子，需要比武，比武對象是已經成為少林寺弟子的人，就是在他之前成為完成比武的人，需要與武力值相近的人比武，如果有多個，那麼選擇比自己武力值低的。輸出自己的id還有與自己比武的人的id.
思路：
用set和map還有lower_bound。set用於去重排序，map用於映射武力值和編號id,lower_bound用於查找與自己武力值相近的和尚。詳細見代碼。
代碼：
1 #include <iostream>
2 #include <set>//利用set可去重按升序排序
3 #include <map>//利用map將id與武力值映射
4 #include <cmath>
5 using namespace std;
6 int main(){
7 int n,k,g;
8 set<int>s;
9 map<int,int>m;
10 while(cin>>n){
11 if(n==0) break;
12 s.clear();
13 m.clear();
14 s.insert(1000000000);//將master入集合
15 m[1000000000]=1;//記錄master的編號
16 while(n--){
17 cin>>k>>g;//輸入新和尚的編號k與武力值g
18 cout<<k<<" ";
19 set<int>::iterator it1,it2;
20 it1=s.lower_bound(g);//it1為第一個武力值大於等於g的和尚的地址
21 if(it1==s.begin()) cout<<m[*it1]<<endl;
22 else {
23 it2=it1--;
24 if(abs(*it1-g)>abs(*it2-g)) cout<<m[*it2]<<endl;
25 else cout<<m[*it1]<<endl;
26 }
27 s.insert(g);//當前新和尚比武完成，將他的武力值入集合，供以後使用
28 m[g]=k;//記錄該武力值g所對應的編號k
29 }
30 }
31 return 0;
32 }
