B. New Year Permutation （CF）-CFANZ编程社区

B. New Year Permutation

time limit per test

memory limit per test

input

output

p1, p2, ..., pn. As the New Year is coming, he wants to make his permutation as pretty as possible.

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

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.

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

Input

n (1 ≤ n ≤ 300) — the size of the permutation p.

n space-separated integers p1, p2, ..., pn — the permutation p that user ainta has. Each integer between 1and n

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 ≤ n, Ai, j = Aj, i holds. Also, for all integers i where 1 ≤ i ≤ n, Ai, i

Output

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

output

1 2 3 4 5

Note

(p1, p7).

(p1, p3), (p4, p5), (p3, p4).

B. New Year Permutation （CF）_ios

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

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>

using namespace std;

int a[310];
char map[301][301];
int v[310];
int p[310];
int n;

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
        for(int i=0;i<n;i++)
        {
            scanf("%s",map[i]);
        }
        memset(v,0,sizeof(v));
        int h;
        for(int i=0;i<n;i++)
        {
            int q[3001],s = 0, e = 0;
            int flag[301];
            memset(flag,0,sizeof(flag));
            q[e++] = i;
            flag[i] = 1;
            while(s<e)
            {
                int x = q[s++];
                for(int j=0;j<n;j++)
                {
                    if(flag[j] == 0 && map[x][j] == '1')
                    {
                        flag[j] = 1;
                        q[e++] = j;
                    }
                }
            }
            int minn = 999999;
            for(int j=0;j<n;j++)
            {
                if(flag[j] == 1 && v[j] == 0)
                {
                    if(a[j] < minn)
                    {
                        minn = a[j];
                        h = j;
                    }
                }
            }
            v[h] = 1;
            p[i] = minn;
        }
        for(int i=0;i<n;i++)
        {
            if(i == 0)
            {
                 printf("%d",p[i]);
            }
            else
            {
                printf(" %d",p[i]);
            }
        }
        printf("\n");
    }
    return 0;
}