11-散列4 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 (≤), 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
Sample Output:
1 13 12 21 33 34 38 27 22 32
//参考
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
const int N = ;
int num[N], indegree[N];
struct cmp {
bool operator()(int i, int j) {
return num[i] > num[j];
}
};
int main() {
int i, j, n, m, k, flag = ;
scanf("%d", &n);
vector<vector<int> > g(n);
priority_queue<int, vector<int>, cmp> q;
for (i = ; i < n; i++)
scanf("%d", &num[i]); for (i = ; i < n; i++) {
if (num[i] > ) {
k = num[i] % n;
indegree[i] = (i + n - k) % n;
if (indegree[i]) {
for (j = ; j <= indegree[i]; j++)
g[(k + j) % n].push_back(i);
}
else q.push(i);
}
} while (!q.empty()) {
i = q.top();
q.pop();
if (!flag) {
flag = ;
printf("%d", num[i]);
}
else printf(" %d", num[i]);
for (j = ; j < g[i].size(); j++) {
if (--indegree[g[i][j]] == )
q.push(g[i][j]);
}
}
return ; }
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