【CodeForces 500B】【贪心】New Year Permutation

New Year Permutation

Time Limit: 2000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u

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 ≤ 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.

Samples

Input1

7

5 2 4 3 6 7 1

0001001

0000000

0000010

1000001

0000000

0010000

1001000

Output1

1 2 4 3 6 7 5

Input2

5

4 2 1 5 3

00100

00011

10010

01101

01010

Output2

1 2 3 4 5

Hint

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.

Source

Good Bye 2014

也是一个比较基本的贪心，就是把矩阵跑一个floyed，然后从前向后枚举每一个数，看每一个数可以和后面的那些交换，如果交换了以后更优就交换，这样一定最后出来字典序是最小的。

实际上感觉是一个图，整个可以分成很多个连通块，然后把每个连通块排序，然后在这样输出来

少说一点了，代码见下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<cstdlib>
#include<cmath>
#include<ctime>
#include<stack>
#define INF 2100000000
#define ll long long
#define clr(x)  memset(x,0,sizeof(x))
#define clrmax(x)  memset(x,127,sizeof(x))

using namespace std;

inline int read()
{
char c;
int ret=0;
while(!(c>='0'&&c<='9'))
c=getchar();
while(c>='0'&&c<='9')
{
ret=(c-'0')+(ret<<1)+(ret<<3);
c=getchar();
}
return ret;
}

#define M 305

int a[M][M],p[M],n;

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

int main()
{
n=read();
for(int i=1;i<=n;i++)
p[i]=read();
for(int i=1;i<=n;i++,getchar())
for(int j=1;j<=n;j++)
{
char c;
scanf("%c",&c);
a[i][j]=c-'0';
}
floyed();
for(int i=1;i<=n;i++)
for(int j=i+1;j<=n;j++)
if(a[i][j]&&p[j]<p[i])
swap(p[i],p[j]);
for(int i=1;i<=n;i++)
printf("%d ",p[i]);
return 0;
}

大概就是这个样子，如果有什么问题，或错误，请在评论区提出，谢谢。