PTA 11-散列4 Hashing - Hard Version

PTA 11-散列4 Hashing - Hard Version

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.

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.

Sample Input:
11
33 1 13 12 34 38 27 22 32 -1 21
结尾无空行

Sample Output:
1 13 12 21 33 34 38 27 22 32
结尾无空行

思路

1.观察出（我是看陈越姥姥的视频）这是个拓扑排序

2.优先队列priority_queue相关知识
参考链接
定义：priority_queue<Type, Container, Functional>

Type 就是数据类型，Container 就是容器类型（Container必须是用数组实现的容器，比如vector,deque等等，但不能用 list。STL里面默认用的是vector），Functional 就是比较的方式，当需要用自定义的数据类型时才需要传入这三个参数，使用基本数据类型时，只需要传入数据类型，默认是大顶堆

默认：

//升序队列
priority_queue <int,vector<int>,greater<int> > q;
//降序队列
priority_queue <int,vector<int>,less<int> >q;

重写函数cmp

struct cmp {
    bool operator()(int i, int j) {
        return Hash[i] > Hash[j];//降序
    }
};

Code

#include<iostream>
#include<vector>
#include<queue>
#define MaxN 1001
using namespace std;
int Hash[MaxN];
struct cmp {
    bool operator()(int i, int j) {
        return Hash[i] > Hash[j];
    }
};
int main(){
    int N;
    scanf("%d",&N);
    vector<vector<int> > v(N);//记录每个对象的拓扑前的对象下标
    vector<int> Indegree(N);//入度的数组下标
    priority_queue<int,vector<int>,cmp> q;//用来输出一样的Indegree值,Hash值最小的数
    for(int i=0;i<N;i++){
        cin>>Hash[i];
    }
    for(int i=0;i<N;i++){
        if(Hash[i]>=0){//非负->不为空
            int key=Hash[i]%N;
            Indegree[i]=(N+i-key)%N;//有多少个数要在这个数之前输出
            if(Indegree[i]==0){//在该有的位置上
                q.push(i);//入队
            }else{
                for(int j=0;j<Indegree[i];j++){
                    v[(key+j)%N].push_back(i);
                }
            }
        }
    }
    int flag=1;
    while(!q.empty()){
        int t=q.top();
        q.pop();
        if(flag){
            flag=0;
            printf("%d",Hash[t]);
        }else{
            printf(" %d",Hash[t]);
        }
        for(int i=0;i<v[t].size();i++){
            if(--Indegree[v[t][i]]==0){
                q.push(v[t][i]);
            }
        }
    }
    printf("\n");
    return 0;
}