Swaps in Permutation
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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, sopk = 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 apositions, 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 61 2 3 4 5 6 7 8 91 44 72 55 83 66 9
7 8 9 4 5 6 1 2 3
给你n个数
下m行,每行两个数,这两个位置的数可以无限次的交换
让你输出一个字典序最大的排序
因为1,4可以无限次的交换,4,7也可以无限次的交换,所以1,7也可以交换,这样可以想到用并查集把他们划为一类,
然后再用优先队列把同一个集合的存进去,这样输出的时候就能保证大的在前面。
#include<queue>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;int f[1000010];int a[1000010];int find(int x){ if(x == f[x]) return f[x]; else return f[x] = find(f[x]);}priority_queue<int> q[1000010];int main(void){ int n,m,i; while(scanf("%d%d",&n,&m)==2) { for(i=0;i<=n;i++) f[i] = i; for(i=1;i<=n;i++) scanf("%d",&a[i]); for(i=1;i<=m;i++) { int t1,t2; scanf("%d%d",&t1,&t2); int x = find(t1); int y = find(t2); if(x != y) f[x] = y; } for(i=1;i<=n;i++) q[f[find(i)]].push(a[i]); for(i=1;i<=n;i++) { printf("%d ",q[f[i]].top()); q[f[i]].pop(); } }}
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