标签：nta contain har 之间 数据结构 sample 最小值 == tin
Given a hash table of size N, we can define a hash function H(x)=x%N. 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.
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.
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.
11
33 1 13 12 34 38 27 22 32 -1 211 13 12 21 33 34 38 27 22 32
给定一个size为n的hashtable ,-1表示该位置没有数据
逆推输入队列
维护一个动态队列, 每次从中取出一个最小值输出, 队列中存放的是可以放入hash列表的值,
其中包括 1.自身哈希值等于列表最终位置 2.哈希值不等于列表最终位置,但是列表最终位置和哈希值之间的元素都被占满了的元素
每趟循环取出可以输出的最小值之后, 重新遍历待处理的剩余列表 , 继续取出可输出元素放入输出队列
#include <iostream> #include <vector> #include <algorithm> #include <unordered_map> using namespace std; int main(){ int n,temp,index; cin >> n; vector<int> outputList;//待输出列表 vector<int> visited(n,0); vector<int> restList;//不能输出的待处理列表 unordered_map<int, int> indexmap;//元素在hash表中的最终位置 for(int i=0;i<n;i++){ cin >> temp; if(temp!=-1){ index=temp%n; indexmap[temp]=i; if(index==i){ outputList.push_back(temp); }else{ restList.push_back(temp); } } } bool flag=false; while(!outputList.empty()){ sort(outputList.begin(), outputList.end()); if(flag){ cout <<" "; }else{ flag=true; } cout << outputList.front(); index=outputList.front()%n; outputList.erase(outputList.begin()); for(int i=index;i<visited.size();i=(i+1)%n){ if(visited[i]==0){ visited[i]=1; break; } } for(int j=0;j<restList.size();j++){ if(restList[j]!=-1){ index=restList[j]%n; bool flag{true}; int realIndex=indexmap[restList[j]]; for(int k=index;k!=realIndex;k=(k+1)%n){ if(visited[k]==0){ flag=false; break; } } if(flag){//可以输出当前元素,加入待输出列表,将元素置为空 outputList.push_back(restList[j]); restList[j]=-1; } } } } return 0; }
数据结构 11-散列4 Hashing - Hard Version (30 分)
标签：nta contain har 之间 数据结构 sample 最小值 == tin
原文地址：https://www.cnblogs.com/ichiha/p/14815466.html
