CF 500B New Year Permutation

B. New Year Permutation

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard output

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.

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 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 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.

Output

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

Sample test(s)

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

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.

题目大意:给你一组数据，和每个位置可已交换的位置，要你输出最小的排列

思路:没有什么特别的思路用一个floyd记录每个位置可以到达的位置然后每次找到最小的值然后与可以交换最前面的位置交换一直处理

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
using namespace std;
#define inf 0xffffff
#define maxn 305

int a[maxn];
bool Map[maxn][maxn];
bool vis[maxn];

int main()
{
int n;
char c;
scanf("%d",&n);
getchar();
for(int i = 0; i < n; i++)
scanf("%d",&a[i]);
getchar();
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++){
scanf("%c",&c);
if(c == '1') Map[i][j] = 1;
else Map[i][j] = 0;
}
getchar();
}
for(int k = 0; k < n; k++)
for(int i = 0; i < n; i++){
if(Map[k][i])
for(int j = 0; j < n; j++)
if(Map[i][k] && Map[k][j]) Map[i][j] = 1;
}
memset(vis,0,sizeof(vis));
while(true){
int v = -1;
for(int i = 0; i < n; i++)
if(!vis[i] && (v == -1||a[v] > a[i])) v = i;
if(v == -1) break;
while(true){
int x = -1;
for(int i = 0; i < v; i++)
if(!vis[i] && Map[v][i] &&a[v] < a[i]) {x = i; break;}
if(x == -1) break;
int c = a[v];
a[v] = a[x];
a[x] = c;
v = x;
}
vis[v] = true;
}
printf("%d",a[0]);
for(int i = 1; i < n; i++)
printf(" %d",a[i]);
putchar('\n');
return 0;
}