【CodeForces】500B - New Year Permutation（思维）

最新推荐文章于 2019-03-30 22:17:00 发布

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这是一个关于CodeForces编程竞赛中的题目，涉及到排列的优化和二进制矩阵操作。用户ainta有一个排列p，他希望通过交换两个不同元素来优化排列，但交换条件受限于一个n×n的二进制矩阵A，只有当A[i][j]=1时，元素pi和pj才能交换。目标是找到能够得到的最漂亮的排列，即字典序最小的排列。解决方案是利用并查集维护能交换的数字集合，并将排序后的较小数字优先放置在前面。

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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 ifAi, 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 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).

$\text{[math]}$

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.

题意：给你一个矩阵，1的位置表示横纵坐标序号位置的数字可以交换，问你最后交换后的序列字典序最小是什么。

题解：通过思考发现能交换的数字具有传递性，比如，（1，2）和（2，3）可以推出（1，3），也就是说，能交换的数字所在的集合里的任意元素都能互相交换，那么我们就用并查集维护。

然后把初始数字sort一遍，取小的数字尽量往前放。

代码如下：

#include <cstdio>
#include <stack>
#include <queue>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
#define CLR(a,b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define LL long long
struct node
{
	int num;
	int pos;
}a[311];
int f[311];
int find(int x)
{
	if (x != f[x])
		f[x] = find(f[x]);
	return f[x];
}
void join(int x,int y)
{
	int fx,fy;
	fx = find(x);
	fy = find(y);
	if (fx != fy)
		f[fx] = fy;
}
bool cmp(node a,node b)
{
	return a.num < b.num;
}
int main()
{
	int n;
	scanf ("%d",&n);
	for (int i = 1 ; i <= n ; i++)
	{
		scanf ("%d",&a[i].num);
		a[i].pos = i;
	}
	for (int i = 1 ; i <= n ; i++)
		f[i] = i;
	getchar();
	for (int i = 1 ; i <= n ; i++)
	{
		for (int j = 1 ; j <= n ; j++)
		{
			char flag;
			scanf ("%c",&flag);
			if (j > i && flag == '1')
				join(i,j);
		}
		getchar();
	}
	int used[311];
	CLR(used,-1);
	sort(a+1,a+1+n,cmp);
	for (int i = 1 ; i <= n ; i++)
	{
		for (int j = 1 ; j <= n ; j++)
		{
			if (used[j] == -1)
			{
				if (find(a[i].pos) == find(j))
				{
					used[j] = a[i].num;
					break;
				}
			}
		}
	}
	for (int i = 1 ; i < n ; i++)
		printf ("%d ",used[i]);
	printf ("%d\n",used[n]);
	return 0;
}