【CodeForces】500B - New Year Permutation（Floyd）（贪心）

New Year Permutation

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 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.

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

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
d83a
of the permutation p is
denoted as pi.
The size of the permutation p is denoted as n.

题意：给定一个序列a[]，再给出关系矩阵g[][]，g[i][j]=1代表a[i]和a[j]可以交换位置，问经过交换能得到字典序最小的序列是多少

思路：可以用Floyd算法来更新是否可以直接或间接的交换（也可以用并查集来维护），如果map[i][k]=1&&map[k][j]=1，则i也可以和j交换，接下来再用贪心的思想，从最小的数开始遍历，判断其是否能往前移动

代码：

#include<stdio.h>
#include<string.h>
#define maxn 300+10
int n;
int a[maxn],map[maxn][maxn],pl[maxn];

void Floyd()
{
for(int k=0; k<n; k++)
for(int i=0; i<n; i++)
for(int j=0; j<n; j++)
if(map[i][k]&&map[k][j])
map[i][j]=1;
}

int main()
{
scanf("%d",&n);
for(int i=0; i<n; i++)
{
scanf("%d",&a[i]);
pl[a[i]]=i;//存储每个数原先的位置
}
char s[maxn];
for(int i=0; i<n; i++)
{
scanf("%s",s);
for(int j=0; j<n; j++)
map[i][j]=s[j]-'0';
}
Floyd();
for(int i=1; i<=n; i++)//从最小的数开始遍历，观察其是否能变到前面
{
for(int j=0; j<n; j++)//从第一个位置开始判断
{
if(map[pl[i]][j]&&i<a[j])//可以交换并且第j位置上的数大于i
{
if(pl[i]>j)//如果数i的位置pl[i]比位置j大，则交换
{
a[pl[i]]=a[j];//将原先i位置上的数变为a[j]
pl[a[j]]=pl[i];//将a[j]的位置更新为pl[i]
a[j]=i;//将位置j上的数变为i
break;
}
}
}
}
for(int i=0;i<n;i++)
printf("%d ",a[i]);
return 0;
}