codeforces500B
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 < bkall 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.
Examples
7
5 2 4 3 6 7 1
0001001
0000000
0000010
1000001
0000000
0010000
1001000
1 2 4 3 6 7 5
5
4 2 1 5 3
00100
00011
10010
01101
01010
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).
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.
sol:十分显然的是a可以与b换,b可以与c换,a就可以和c换,于是把可以换的用并查集缩成一个个连通块,在每个联通块中做到最优,那么就是答案了
#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline ll read()
{
ll s=;
bool f=;
char ch=' ';
while(!isdigit(ch))
{
f|=(ch=='-'); ch=getchar();
}
while(isdigit(ch))
{
s=(s<<)+(s<<)+(ch^); ch=getchar();
}
return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
if(x<)
{
putchar('-'); x=-x;
}
if(x<)
{
putchar(x+''); return;
}
write(x/);
putchar((x%)+'');
return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
const int N=;
int n,a[N],Pos[N];
bool Bo[N][N];
int Father[N];
inline int Get_Father(int x)
{
return (Father[x]==x)?(x):(Father[x]=Get_Father(Father[x]));
}
struct Record
{
int Shuz,Weiz;
};
inline bool cmp_Shuz(Record p,Record q)
{
return p.Shuz<q.Shuz;
}
inline bool cmp_Weiz(Record p,Record q)
{
return p.Weiz<q.Weiz;
}
vector<Record>Liantong[N];
vector<int>Xulie[N];
int main()
{
int i,j;
R(n);
for(i=;i<=n;i++)
{
R(a[i]); Father[i]=i;
}
for(i=;i<=n;i++)
{
for(j=;j<=n;j++)
{
char ch=' ';
while(!isdigit(ch)) ch=getchar();
if(ch=='')
{
Father[Get_Father(i)]=Get_Father(j);
}
}
}
for(i=;i<=n;i++)
{
Liantong[Get_Father(i)].push_back((Record){a[i],i});
}
for(i=;i<=n;i++) if(Liantong[i].size())
{
sort(Liantong[i].begin(),Liantong[i].end(),cmp_Shuz);
for(j=;j<Liantong[i].size();j++) Xulie[i].push_back(Liantong[i][j].Shuz);
}
for(i=;i<=n;i++) if(Liantong[i].size())
{
sort(Liantong[i].begin(),Liantong[i].end(),cmp_Weiz);
for(j=;j<Liantong[i].size();j++) a[Liantong[i][j].Weiz]=Xulie[i][j];
}
for(i=;i<n;i++) W(a[i]); Wl(a[n]);
return ;
}
/*
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
*/
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