HDU 4585 Shaolin

Shaolin

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 195    Accepted Submission(s): 99


Problem DescriptionShaolin 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. 

InputThere 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. 

OutputA 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

 

Source2013ACM-ICPC杭州賽區全國邀請賽 題意： 有N個依次進入少林， 每次輸出， 新進和尚 和 戰鬥等級與其最接近的舊和尚的ID。  ID ， 和 戰鬥等級都是唯一的。思路： STL的應用

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <vector>#include <map>#include <set>#include <cmath>using namespace std;//const int V = 40 + 5;const int MaxN = 80 + 5;const int mod = 10000 + 7;const __int64 INF = 0x7FFFFFFFFFFFFFFFLL;const int inf = 0x7fffffff;map<int, int> m;int n;int main() {    int i, j;    while(scanf("%d", &n), n) {        m[1000000000] = 1;        for(i = 0; i < n; ++i) {            int id, fg, ans;            scanf("%d%d", &id, &fg);            m[fg] = id;            map<int, int>::iterator it = m.find(fg);            if(it == m.begin())                ans = (++it)->second;            else {                map<int, int>::iterator it1 = it;                if(abs((++it1)->first - fg) >= abs((--it)->first - fg))                    ans = it->second;                else                    ans = it1->second;            }            printf("%d %d\n", id, ans);        }        m.clear();    }}