cf 500B

B - New Year Permutation
Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u
Appoint description: 

Description

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 ≤ ni ≠ j) if and only ifAi, j = 1.

Given the permutation p and the matrix A, user ainta wants to know the prettiest permutation that he can obtain.

Input

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 andn occurs 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 thei-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 ≤ nAi, j = Aj, i holds. Also, for all integers i where 1 ≤ i ≤ nAi, i = 0 holds.

Output

In the first and only line, print n space-separated integers, describing the prettiest permutation that can be obtained.

Sample Input

Input
7
5 2 4 3 6 7 1
0001001
0000000
0000010
1000001
0000000
0010000
1001000
Output
1 2 4 3 6 7 5
Input
5
4 2 1 5 3
00100
00011
10010
01101
01010
Output
1 2 3 4 5

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#include<queue>
#include<vector>
#include<set>
using namespace std;
int a[100010],n;
char s[410][410];
int main()
{
     scanf("%d",&n);
     for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
     for(int i=0;i<n;i++)
      scanf("%s",s[i]);
     for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
                  for(int k=0;k<n;k++)
                        if(s[j][i]=='1'&&s[i][k]=='1')
                              s[j][k]='1';
     for(int i=0;i<n;i++)
            for(int j=i+1;j<n;j++)
                  if(s[i][j]=='1'&&a[i]>a[j])
                        swap(a[i],a[j]);
     for(int i=0;i<n-1;i++)
            printf("%d ",a[i]);
     printf("%d\n",a[n-1]);
     return 0;
}

  

posted @ 2015-01-12 09:01  心田定则  阅读(121)  评论(0编辑  收藏  举报