Algorithm
You are given a permutation of the numbers 1, 2, ..., n and m pairs of positions (aj, bj).
At each step you can choose a pair from the given positions and swap the numbers in that positions. What is the lexicographically maximal permutation one can get?
Let p and q be two permutations of the numbers 1, 2, ..., n. p is lexicographically smaller than the q if a number 1 ≤ i ≤ n exists, so pk = qk for 1 ≤ k < i and pi < qi.
The first line contains two integers n and m (1 ≤ n, m ≤ 106) — the length of the permutation p and the number of pairs of positions.
The second line contains n distinct integers pi (1 ≤ pi ≤ n) — the elements of the permutation p.
Each of the last m lines contains two integers (aj, bj) (1 ≤ aj, bj ≤ n) — the pairs of positions to swap. Note that you are given a positions, not the values to swap.
Print the only line with n distinct integers p'i (1 ≤ p'i ≤ n) — the lexicographically maximal permutation one can get.
9 6
1 2 3 4 5 6 7 8 9
1 4
4 7
2 5
5 8
3 6
6 9
7 8 9 4 5 6 1 2 3
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int const N = 1e6 + 1;
int n, m;
int nums[N], ds[N], sols[N];
vector<vector<int> > vals, poss;
int find(int x) {
return (ds[x] == x ? x : (ds[x] = find(ds[x])));
}
int main() {
cin >> n >> m;
for(int i = 0; i < n; i++) {
ds[i] = i;
cin >> nums[i];
}
int a, b;
for(int i = 0; i < m; i++) {
cin >> a >> b;
a--; b--;
ds[find(a)] = find(b);
}
vals.resize(n);
poss.resize(n);
for(int i = 0; i < n; i++) {
vals[find(i)].push_back(nums[i]);
poss[find(i)].push_back(i);
}
for(int i = 0; i < n; i++) {
sort(vals[i].rbegin(), vals[i].rend());
for(int j = 0; j < poss[i].size(); j++)
sols[poss[i][j]] = vals[i][j];
}
for(int i = 0; i < n; i++)
cout << (i == 0 ? "" : " ") << sols[i];
cout << endl;
return 0;
}Input
1 2 3 4 5 6 7 8 9
1 4
4 7
2 5
5 8
3 6
6 9
Output
Demonstration
Codeforces Solution-Swaps in Permutation-Solution in C, C++, Java, Python