Good Bye 2014-B. New Year Permutation

Good Bye 2014--B. 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 1and 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 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 = 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).

$Good Bye 2014-B. New Year Permutation$

A 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.

题意：给n个数a[]，然后给n*n的关系矩阵g[][]，如果关系矩阵中a[i][j] == 1，则代表a[i]和a[j]等价，可以交换。问经过交换能得到的字典序最小的序列为多少。

解析：利用Floyd算法，把所有可以直接或间接交换的的关系矩阵都赋值为1，这样再暴力贪心是能交换的尽可能把小的元素交换到尽可能靠前的位置即可。

AC代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
using namespace std;
#define INF 0x7fffffff
#define LL long long
#define MID(a, b)  a+(b-a)/2
const int maxn = 1000 + 10;

#pragma comment(linker, "/STACK:1024000000,1024000000")

template <class T>
inline bool scan_d(T &ret) {
    char c; int sgn;
    if(c=getchar(),c==EOF) return 0; //EOF
    while(c!=' -' &&(c<'0' ||c>'9' )) c=getchar();
    sgn=(c==' -' )?-1:1;
    ret=(c==' -' )?0:(c-'0' );
    while(c=getchar(),c>='0' &&c<='9' ) ret=ret*10+(c-'0' );
    ret*=sgn;
    return 1;
}

inline void out(int x) {
    if(x>9) out(x/10);
    putchar(x%10+'0' );
}

//int pa[305];
int a[305];
char s[305][305];
int b[305][305];


int main(){
    #ifdef sxk
        freopen("in.txt", "r", stdin);
    #endif // sxk

    std::ios::sync_with_stdio(false);
    std::cin.tie(0);

    int n;
    while(scanf("%d", &n)!=EOF){
        for(int i=1; i<=n; i++) scanf("%d", &a[i]);
        for(int i=1; i<=n; i++) scanf("%s", &s[i]);
        for(int i=1; i<=n; i++){
            for(int j=0; j<n; j++){
                if(s[i][j] == '1') b[i][j+1] = 1;
                else b[i][j+1] = 0;
            }
        }
        for(int k=1; k<=n; k++){
            for(int i=1; i<=n; i++){
                for(int j=1; j<=n; j++){
                    if(b[i][k] && b[k][j]) b[i][j] = 1;
                }
            }
        }
        for(int i=1; i<=n; i++)
            for(int j=i+1; j<=n; j++)
                if(b[i][j] && a[i] > a[j])
                    swap(a[i], a[j]);
        for(int i=1; i<=n; i++)
            printf("%d%c", a[i], i < n ? ' ' : '\n');
    }
    return 0;
}

Good Bye 2014-B. New Year Permutation

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