CodeForces - 500B 最短路模拟暴力

User ainta has a permutation p₁, p₂, ..., p_n. As the New Year is coming, he wants to make his permutation as pretty as possible.

Permutation a₁, a₂, ..., a_n is prettier than permutation b₁, b₂, ..., b_n, if and only if there exists an integer k (1 ≤ k ≤ n) where a₁ = b₁, a₂ = b₂, ..., a_k - 1 = b_k - 1 and a_k < b_k 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 ann × n binary matrix A, user ainta can swap the values of p_i and p_j (1 ≤ i, j ≤ n, i ≠ j) if and only if A_i, 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, ..., p_n — the permutationp that user ainta has. Each integer between 1 and n 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 the i-th line A_i, 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, A_i, j = A_j, i holds. Also, for all integers i where 1 ≤ i ≤ n, A_i, i = 0 holds.

Output

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

Examples

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

Output

1 2 3 4 5

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).

$\text{[math]}$

A permutation p is a sequence of integers p₁, p₂, ..., p_n, consisting of n distinct positive integers, each of them doesn't exceed n. The i-th element of the permutation p is denoted as p_i. The size of the permutation p is denoted as n.

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题意：给n个数的数组num[]，给出关系矩阵，d[i][j]=1表示可以交换num[i]，num[j];求经过有限次交换后字典序最小的num[];

思路: 用一个p数组保存num[i]的位置，如果数i的位置p[i]可以和位置j交换,并且数i的位置大于位置j,并且i小于num[j](位置j对应的数)，则交换num[i],num[j];

#include<bits\stdc++.h>
using namespace std;
#define  ll long long
#define  ull unsigned long long
#define mINF(a) memset(a,0x3f3f3f3f,sizeof(a))
#define m0(a) memset(a,0,sizeof(a))
#define mf1(a) memset(a,-1,sizeof(a))
const int INF=0x3f3f3f3f;
const int len=3e2+5;
const ll MAX=1e18;
int num[len];
int n;
int d[len][len];
void floyd()
{
	for(int k=0;k<n;++k)
		for(int i=0;i<n;++i)
			for(int j=0;j<n;++j)
				if(d[i][k]&&d[k][j])
					d[i][j]=1;
}
int main()
{
	while(cin>>n)
	{
		int p[len]={};
		for(int i=0;i<n;++i)
		{
			scanf("%d",num+i);
			p[num[i]]=i;//num[i]的位置为p[num[i]]; 
		}
		for(int i=0;i<n;++i)
			for(int j=0;j<n;++j)
				scanf("%1d",&d[i][j]);//在紫书上看的,"%nd",输入一行数中的前n位 
		floyd();//最短路的算法,本题的应用好像叫传递闭包; 
		for(int i=1;i<=n;++i)//i从小到大枚举, 
		{
			for(int j=0;j<n;++j)//枚举num的位置 
			{
				if(d[p[i]][j]&&p[i]>j&&i<num[j])//如果数i的位置p[i]可以和位置j交换,并且数i的位置大于位置j,并且i小于num[j](位置j对应的数) 
				{
					num[p[i]]=num[j];//将位置j的数赋给i所在位置的数 
					p[num[j]]=p[i];//将数i的位置赋给位置j对应数的位置 
					num[j]=i;//将数i赋给位置j对应的数
					//不明白模拟一下就好
					//可以不更新数i最终的位置 
					break;
				}
			}
		}
		for(int i=0;i<n;++i)
			printf("%d ",num[i]);
	}
}

CodeForces - 500B 最短路 模拟 暴力

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CodeForces - 500B 最短路模拟暴力