7-18 Hashing - Hard Version

7-18 Hashing - Hard Version （30 分）

Given a hash table of size N, we can define a hash function . Suppose that the linear probing is used to solve collisions, we can easily obtain the status of the hash table with a given sequence of input numbers.

However, now you are asked to solve the reversed problem: reconstruct the input sequence from the given status of the hash table. Whenever there are multiple choices, the smallest number is always taken.

Input Specification:

Each input file contains one test case. For each test case, the first line contains a positive integer N (≤1000), which is the size of the hash table. The next line contains N integers, separated by a space. A negative integer represents an empty cell in the hash table. It is guaranteed that all the non-negative integers are distinct in the table.

Output Specification:

For each test case, print a line that contains the input sequence, with the numbers separated by a space. Notice that there must be no extra space at the end of each line.

 

知识点：

拓扑排序

priority_queue的用法：

• priority_queue<int,vector<int>,greater<int> > q; 建立小顶堆

• priority_queue<int,vector<int>,less<int> > q; 建立大顶堆

map 的一些用法：

• 遍历一个 map  
    for(map<int,int>::iterator it=mp.begin();it!=mp.end();it++){
          if(list[it->second]==0){
              q.push(it->first);
          }
      }

   

思路：

这是一个拓扑排序问题。建立一个优先队列（最小堆）。寻找所有元素，将入度为0的入队，然后将所有以它为前置节点的项，入度都减1。

入度的计算：考虑到是线性探测，每个元素的入度就是（它目前在的位置 - 应该在的位置），如负，加hash表长。

用 map 建立每个数的索引，不然空间不够。

#include <iostream>
#include <cmath>
#include <queue>
#include <vector>
#include <map>
using namespace std;
const int maxn = 1005;
int n;
int a[maxn];
vector<int> zu[maxn];
priority_queue<int,vector<int>,greater<int> > q;
int list[maxn];
map<int,int> mp;

int main(){
    scanf("%d",&n);
    fill(list,list+maxn,-1);
    for(int i=0;i<n;i++){
        scanf("%d",&a[i]);
        if(a[i]<=-1) continue;
        mp[a[i]] = i;
        list[i] = i-a[i]%n;
        if(list[i]<0) list[i]+=n;    
    }
    for(int i=0;i<n;i++){
        if(a[i]<=-1) continue;
        int j=a[i]%n;
        while(j!=i){
            zu[mp[a[j]]].push_back(a[i]);
            j+=1;
            if(j>=n) j-=n;
        }
    }
        for(map<int,int>::iterator it=mp.begin();it!=mp.end();it++){
        if(list[it->second]==0){
            q.push(it->first);
        }
    }
    //printf("12 %d\n",list[12]);
    vector<int> out;
    while(q.size()){
        int tmp = q.top();
        q.pop();
        out.push_back(tmp);
        for(int i=0;i<zu[mp[tmp]].size();i++){
            list[mp[zu[mp[tmp]][i]]]--;
            //printf("32: %d\n",list[mp[32]]);
            if(list[mp[zu[mp[tmp]][i]]]==0){
                q.push(zu[mp[tmp]][i]);
            }
        }
    }
    for(int i=0;i<out.size();i++){
        if(i!=0) printf(" ");
        printf("%d",out[i]);
    }
}

 

Sample Input:

11
33 1 13 12 34 38 27 22 32 -1 21

Sample Output:

1 13 12 21 33 34 38 27 22 32