【CodeForces 500B】【貪心】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

也是一個比較基本的貪心，就是把矩陣跑一個floyed，然後從前向後枚舉每一個數，看每一個數可以和後面的那些交換，如果交換了以後更優就交換，這樣一定最後出來字典序是最小的。
實際上感覺是一個圖，整個可以分成很多個連通塊，然後把每個連通塊排序，然後在這樣輸出來
少說一點了，代碼見下：

#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;
}

大概就是這個樣子，如果有什麼問題，或錯誤，請在評論區提出，謝謝。