cf （dfs，floyd，并查集）. New Year Book Reading

B - New Year Permutation

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 andn 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 Input

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

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. Thei-th element of the permutation p is
denoted as pi. The size of the permutation p is denoted as n.

题意：给出一个序列p和一个矩阵a，问能否找到一个更完美的序列，完美意味着按照最小字典序排序，

序列改变方式为：矩阵为1的数所在的ai。aj所对应的pi，pj可互换。
题解：dfs，floyd ，并查集，找到所有可交换的数，重新排序。

dfs

<span style="font-size:18px;">//题意：给出一个序列p和一个矩阵a，问能否找到一个更完美的序列，
//   完美意味着按照最小字典序排序，
//序列改变方式为：矩阵为1的数所在的ai。aj所对应的pi，pj可互换。
#include<cstdio>
#include<algorithm>
using namespace std;

int n,k,x;
int p[1010],b[1010],c[1010];
char a[1010][1010];

void dfs(int i){
if (!b[p[i]])
x=min(x,p[i]);
c[i]=k;
int j;
for (j=0;j<n;j++){
if (a[i][j]=='1' && c[j]!=k)
dfs(j);
}

}
int main(){
scanf ("%d",&n);
int i;
for (i=0;i<n;i++)
scanf ("%d",&p[i]);
for (i=0;i<n;i++)
scanf ("%s",a[i]);
for (i=0;i<n;i++){
k++;
x=n;
dfs(i);
b[x]=true;
printf ("%d ",x);
}
return 0;
} </span>

floyd

<span style="font-size:18px;">#include<cstdio>
using namespace std;
int n,a[310],b[310][310];
char s[310];
int main()
{
scanf("%d",&n);
int i,j,k;
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
for(i=0;i<n;i++)
{
scanf("%s",s);
for(j=0;j<n;j++)
{
if(s[j]=='1') b[i][j]=1;
else b[i][j]=0;
}
}

}

for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
if(b[i][j]==0) continue;
for(k=0;k<n;k++)
{
if(b[i][k]==1) b[j][k]=1;
}
}
}
for(i=0;i<n;i++)
{
for(j=i+1;j<n;j++)
{
if(b[i][j]==1&&a[i]>a[j])
{
int b=a[i];
a[i]=a[j];
a[j]=b;
}
}
}
for(i=0;i<n;i++)
{
printf("%d",a[i]);
if(i==n-1) printf("\n");
else printf(" ");
}
return 0;
}

</span>

并查集

#include<cstdio>
#include<algorithm>
using namespace std;
int n,p[330],fa[330];
char a[330][330];

int find(int x){
if (x!=fa[x])
fa[x]=find(fa[x]);
return fa[x];
}

void Union(int a,int b){
int fx=find(a);
int fy=find(b);
if (fx!=fy)
fa[fy]=fx;

}
int main(){
int i,j;
scanf ("%d",&n);
for (i=1; i<=n; i++){
scanf ("%d",&p[i]);
fa[i]=i;
}

for (i=1; i<=n; i++){
scanf ("%s",a[i]);
for (j=0; j<n; j++){ //注意j
if (a[i][j]=='1')
Union(i,j+1);
}
}
for (i=1;i<=n;i++){
int q=i;
for (j=i+1; j<=n; j++)
if (p[q]>p[j]&&(find(i)==find(j)))
q=j;
swap(p[q],p[i]);
printf ("%d ",p[i]);
}

return 0;
}