Good Bye 2014 B. New Year Permutation 并查集 最短路 floyed算法

B. New Year Permutation

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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)
wherea1 = 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 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 where1 ≤ 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).



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，给出一个初始数列，再给出一个矩阵，确定i j位置是否可以交换。要求最小字典序的数列。

首先给一个简单的想法，如果知道任意两点是否可以交换（包括间接交换），那么，我们只需要从最小的数开始一个一个的确定最小的位置 ，就可以得出答案了。如果要知道间接是否可交换，相当于求，任意两点间，最短路是否可为1，用floyed算法求出，复杂度为o(n ^3)

#define N 305
#define M 100005
#define maxn 205
#define MOD 1000000000000000007
int n,b

,p
,a
;
char str
;
void prin(){
FI(n)
if(!i)
printf("%d",p[i] + 1);
else
printf(" %d",p[i] + 1);
printf("\n");
}
int main()
{
//freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
while(S(n)!=EOF)
{
FI(n) {
S(p[i]);
p[i]--;
a[p[i]] = i;
}
FI(n) {
SS(str);
for(int j = 0;str[j] != '\0';j++){
b[i][j] = str[j] == '1'?1:0;
}
}
for(int i = 0;i<n;i++)
for(int j = 0;j<n;j++)
for(int k = 0;k<n;k++){
b[j][k] = max(b[j][k],b[j][i] & b[i][k]);
}
FI(n){
for(int j = 0;j<a[i];j++){
if(p[j] > i && b[a[i]][j]){
int t = p[j],t2 = a[i];
p[j] = i;
p[t2] = t;
a[i] = j;
a[t] = t2;
break;
}
}

}
prin();
}
//fclose(stdin);
//fclose(stdout);
return 0;
}

第二种方法。用并查集优化，由于，在同一个集合内的点，明显，只需要从小到大排（因为同一集合，经过一点变换，一定可以生成各种排列），就一定是最优解了。复杂度，为O(n * n);
#define N 305
#define M 100005
#define maxn 205
#define MOD 1000000000000000007
int n,p
,num
;
char str
;
vector<int> g
;
struct UnionFind{
int fa
;
int findFa(int x){
return fa[x] == x? x:fa[x] = findFa(fa[x]);
}
void init(int n){
FI(n) fa[i] = i;
}
void Uni(int a,int b){
fa[findFa(a)] = findFa(b);
}
}U;
int main()
{
//freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
while(S(n)!=EOF)
{
U.init(n);
fill(num,0);
FI(n) {
S(p[i]);
p[i]--;
}
FI(n) {
SS(str);
for(int j = 0;str[j] != '\0';j++){
if(str[j] == '1'){
U.Uni(i,j);
}
}
}
FI(n) g[U.findFa(i)].push_back(p[i]);
FI(n) if(U.findFa(i) == i) sort(g[i].begin(),g[i].end());
FI(n){
int fa = U.findFa(i);
int t = g[fa][num[fa]++];
if(!i) printf("%d",t + 1);
else printf(" %d",t + 1);
}
printf("\n");
}
//fclose(stdin);
//fclose(stdout);
return 0;
}