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。
1 #include <iostream>
2 #include <cmath>
3 #include <queue>
4 #include <vector>
5 #include <map>
6 using namespace std;
7 const int maxn = 1005;
8 int n;
9 int a[maxn];
10 vector<int> zu[maxn];
11 priority_queue<int,vector<int>,greater<int> > q;
12 int list[maxn];
13 map<int,int> mp;
14
15 int main(){
16 scanf("%d",&n);
17 fill(list,list+maxn,-1);
18 for(int i=0;i<n;i++){
19 scanf("%d",&a[i]);
20 if(a[i]<=-1) continue;
21 mp[a[i]] = i;
22 list[i] = i-a[i]%n;
23 if(list[i]<0) list[i]+=n;
24 }
25 for(int i=0;i<n;i++){
26 if(a[i]<=-1) continue;
27 int j=a[i]%n;
28 while(j!=i){
29 zu[mp[a[j]]].push_back(a[i]);
30 j+=1;
31 if(j>=n) j-=n;
32 }
33 }
34 for(map<int,int>::iterator it=mp.begin();it!=mp.end();it++){
35 if(list[it->second]==0){
36 q.push(it->first);
37 }
38 }
39 //printf("12 %d
",list[12]);
40 vector<int> out;
41 while(q.size()){
42 int tmp = q.top();
43 q.pop();
44 out.push_back(tmp);
45 for(int i=0;i<zu[mp[tmp]].size();i++){
46 list[mp[zu[mp[tmp]][i]]]--;
47 //printf("32: %d
",list[mp[32]]);
48 if(list[mp[zu[mp[tmp]][i]]]==0){
49 q.push(zu[mp[tmp]][i]);
50 }
51 }
52 }
53 for(int i=0;i<out.size();i++){
54 if(i!=0) printf(" ");
55 printf("%d",out[i]);
56 }
57 }
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