HDU 4585 Shaolin（STL map）

Shaolin

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d
& %I64u

Submit 

cid=57992#status//C/0" class="ui-button ui-widget ui-state-default ui-corner-all ui-button-text-only" style="font-family:Verdana,Arial,sans-serif; font-size:1em; border:1px solid rgb(211,211,211); background-color:rgb(227,228,248); color:rgb(85,85,85); display:inline-block; position:relative; padding:0px; margin-right:0.1em; zoom:1; overflow:visible; text-decoration:none">Status Practice HDU
4585

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 

大致题意：非常久曾经。少林寺里仅仅有一个等级10^9为方丈，他编号为1，少林寺里有个规矩，每一个新来的和尚都必须和一位等级最接近自己的老（就是比他先来的）和尚进行战斗。后来又来了n个和尚。而且他们按入寺时间依次编号2~n+1，可是时间太久。方丈已经不知道每一个和尚来的时候是跟哪个和尚战斗了，可是他还是记得哪个和尚比哪个和尚先来的。

输入给出n个和尚的编号和他们的等级。

以下就靠你了，让你依次求出每一个和尚进寺是是跟谁战斗的，并输出。

解题思路：直接开数组的话，每次找跟他等级最接近的肯定要排序。眼下知道的最快的快排的也是O(n*lg n)的，因为数据给的是10^5，再加上还要操作，时间已经超了。

所以，得换个思路，那就得用到STL中的map的优势了，也好久没用map做题了。map里的数据会自己主动按键值升序排列。据说map内部的排序是用红黑树（一种不明觉厉的数据结构）实现的。其时间复杂度是O（n）的。同一时候，map里还能够直接用find（）按键值查找键值所在的迭代器，这就太方便了。

详细实现步骤：1.先开个map<int, int> monk,  把方丈的编号id和等级grad在map里建立映射。这里用到了一个小细节。

建立映射能够直接这样写。monk[grad] = id.

2.在循环输入n个和尚的信息，

1）每次输入一个和尚的信息。然后在map里建立映射；

2）此时用find（grad）寻找当前增加的和尚所在的迭代器。

3）然后，考虑当前的迭代器的位置，若是最后一个。就直接输出当前迭代器的前一个；

要是第一个。就输出当前迭代器的后面一个；

要既不是第一个也不是最后一个的话。就比較当前迭代器的first和它前后两迭代器first之间的差值哪个更小，小的输出；

要是两差值一样的话，输出前一个迭代器。

AC代码：

#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>
#include <time.h>
using namespace std;
#define INF 0x7fffffff

map<int, int> a;      //声明一个map对象

int main()
{
    #ifdef sxk
        freopen("in.txt","r",stdin);
    #endif
    int n, k, g;
    while(scanf("%d",&n)!=EOF && n)
    {
        a[1e9] = 1;       //处理方丈
        for(int i=0; i<n; i++){
            scanf("%d%d", &k, &g);
            a[g] = k;
            map<int, int>::iterator temp = a.find(g);      //按grad查找当前和尚所在的迭代器
            if(temp == a.begin())  printf("%d %d\n", k, (++temp)->second);
            else if(temp == a.end())  printf("%d %d\n", k, (--temp)->second);
            else{
                if(abs((--temp)->first - (++temp)->first) > abs((++temp)->first - (--temp)->first))
                    printf("%d %d\n", k, (++temp)->second);
                else  printf("%d %d\n", k, (--temp)->second);
            }
        }
        a.clear();
    }
    return 0;
}