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.
The first line contains an integer n (1 ≤ n ≤ 300) — the size of the permutation p.
The second line contains n space-separated integers p1, p2, ..., pn — the permutation p that user ainta has. Each integer between 1 and noccurs exactly once in the given permutation.
Next n lines describe the matrix A. The i-th line contains n characters '0' or '1' and describes the i-th row of A. The j-th character of the i-th line Ai, j is the element on the intersection of the i-th row and the j-th column of A. It is guaranteed that, for all integers i, j where 1 ≤ i < j ≤ n, Ai, j = Aj, i holds. Also, for all integers i where 1 ≤ i ≤ n, Ai, i = 0 holds.
In the first and only line, print n space-separated integers, describing the prettiest permutation that can be obtained.
7 5 2 4 3 6 7 1 0001001 0000000 0000010 1000001 0000000 0010000 1001000
1 2 4 3 6 7 5
5 4 2 1 5 3 00100 00011 10010 01101 01010
1 2 3 4 5
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.
题目大意：
给你 n 个数字的序列，然后给你 n 行字符串，如果第 i 行字符串对应的第 j 列的字符是1，说明序列 i 和 j 对应的数字可以互换，求所能互换的最小字典序的序列
解题思路：
并查集，输入每行的时候，如果对应的列是1，就把该行和该列用并查集连接起来，然后再遍历序列的每一个数字，把属于一个联通块的数字丢到一个vector里，然后对每一个vector里面的数字进行排序，最后依次输出每个联通块的数字，用cnt数组记录第 n 个联通块输出到第几个了，搞定。
#include<iostream>
#include<vector>
#include<algorithm>
#include<cstdio>
using namespace std;
int pre[305];
int find(int x){
if(x==pre[x])return x;
else return pre[x]=find(pre[x]);
}
void join(int a,int b){
int ra=find(a);
int rb=find(b);
if(ra!=rb)pre[ra]=rb;
}
int main(){
int n,cnt[305]={0};
int p[305];char s[305];
vector<int> pos[305];
cin>>n;
for(int i = 1; i <= n; i++)pre[i]=i;
for(int i = 1; i <= n; i++)cin>>p[i];
for(int i=1;i<=n;i++){
cin>>s+1;
for(int j=1;j<=n;j++)if(s[j]=='1')join(i,j);
}
for(int i = 1; i <= n; i++)pos[find(i)].push_back(p[i]);
for(int i = 1; i <= n; i++)sort(pos[i].begin(),pos[i].end());
for(int i = 1; i <= n; i++){
int g=find(i);
printf("%d%c",pos[g][cnt[g]++],i==n?'\n':' ');
}
}