5-18 Hashing

5-18 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 $N$ ($\le 1000$), 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.

Sample Input:

1133 1 13 12 34 38 27 22 32 -1 21

Sample Output:

1 13 12 21 33 34 38 27 22 32

参考了其他人的代码，发现他们都用了STL，之前我从来没有用过。用了才知道STL真的方便，这道题可以用STL的优先队列，免去了自己写堆的过程了。

这是数据结构课里最后一题了，也断断续续把数据结构学完了，感觉自己有很多东西都不知道，但也学到了很多东西。

学习的过程中写不出来，没弄懂的感觉太难受了，有时候也怪自己偷懒了。但大部分还是一点一点理解弄懂了，做出来题的时候是最开心的时候了。

以后的学习中还是要坚持，再难都要啃下来！

编程这东西还是要靠自己多看多写多练啊！

#include<stdio.h>  #include<stdlib.h> #include<math.h>#include<string.h>#include<algorithm>#include <vector>  #include <queue>  #include<iostream>  #include <functional> #define MAX 1001#define MAXD 6#define TELNUM 10using namespace std;struct HashNode{int hash;int key;friend bool operator < (HashNode a,HashNode b)/*结构体的优先级设置*/{return a.key > b.key;}};typedef struct HashNode *pHashNode;int Indegree[MAX];void TopSort(vector<vector<int> > rating,int N,pHashNode H){vector<int>TopNum;int i;priority_queue<HashNode> Q;/*利用优先队列*/for(i = 0; i < N; i++){if(!Indegree[i]){Q.push(H[i]);}}while(!Q.empty()){HashNode h = Q.top();int V = h.hash;TopNum.push_back(h.key);Q.pop();for(int i = 0; i < rating[V].size(); i++){if(--Indegree[rating[V][i]] == 0){Q.push(H[rating[V][i]]);}}}printf("%d",TopNum[0]);for(int i = 1 ; i < TopNum.size(); i++){printf(" %d",TopNum[i]);}}int main(void){int N;int i,j,hashpos,currentpos;scanf("%d",&N);vector<vector<int> > rating(N);/*模拟邻接表*/pHashNode H = (pHashNode)malloc(N * sizeof(struct HashNode));for(i = 0; i < N; i++){scanf("%d",&H[i].key);H[i].hash = i;/*每个元素对应一个下标*/if(H[i].key == -1){Indegree[i] = -1;continue;}currentpos = i;hashpos = H[i].key % N;Indegree[i] = (currentpos - hashpos + N) % N;/*计算发生冲突的次数，相当与一个顶点的入度*/for(j = 0; j < Indegree[i]; j++)/*找到每个冲突的地方（指向H[i].key元素的那个元素），把i接入指向每个发生冲突的地方*/{rating[(hashpos + j + N) % N].push_back(i);/*rating[X]代表表头，它里面的元素代表相连的元素*/}}TopSort(rating,N,H);return 0;}