作者：jinyan胡_435 | 来源：互联网 | 2023-10-11 17:06
User ainta has a permutation p1, p2, ..., pn. As the New Year is coming, he wants to make his permutation as pretty as possible.
Permutation a1, a2, ..., an is prettier than permutation b1, b2, ..., bn, if and only if there exists an integer k (1 ≤ k ≤ n) where a1 = b1, a2 = b2, ..., ak - 1 = bk - 1 and ak < bk all holds.
As known, permutation p is so sensitive that it could be only modified by swapping two distinct elements. But swapping two elements is harder than you think. Given an n × n binary matrix A, user ainta can swap the values of pi and pj (1 ≤ i, j ≤ n, i ≠ j) if and only if Ai, j = 1.
Given the permutation p and the matrix A, user ainta wants to know the prettiest permutation that he can obtain.
Output
In the first and only line, print n space-separated integers, describing the prettiest permutation that can be obtained.
Note
In the first sample, the swap needed to obtain the prettiest permutation is: (p1, p7).
In the second sample, the swaps needed to obtain the prettiest permutation is (p1, p3), (p4, p5), (p3, p4).
A permutation p is a sequence of integers p1, p2, ..., pn, consisting of n distinct positive integers, each of them doesn't exceed n. The i-th element of the permutation p is denoted as pi. The size of the permutation p is denoted as n.
位置的swap关系可以构成一个连通块
每一个连通块贪心就行了
实现上，可以用并查集维护，然后对于1到n每一个位置找连通块中最小的
//
// main.cpp
// cf500b
//
// Created by Candy on 9/16/16.
// Copyright © 2016 Candy. All rights reserved.
//
#include
#include
#include
#include
using namespace std;
const int N=305;
int n,p[N],a[N][N],ans[N];
char tmp[N];
int fa[N];
int find(int x){return x==fa[x]?x:fa[x]=find(fa[x]);}
int main(int argc, const char * argv[]) {
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d",&p[i]),fa[i]=i;
for(int i=1;i<=n;i++){
scanf("%s",tmp+1);
for(int j=1;j<=n;j++)
if(tmp[j]=='1'){
int x=find(i),y=find(j);
if(x!=y) fa[y]=x;
}
}
for(int i=1;i<=n;i++){
int mn=i;
for(int j=i+1;j<=n;j++)
if(find(i)==find(j)&&p[j]j;
swap(p[mn],p[i]);
printf("%d ",p[i]);
}
return 0;
}
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