Good Bye 2014 B. New Year Permutation 並查集 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.

題意，給出一個初始數列，再給出一個矩陣，確定i j位置是否可以交換。要求最小字典序的數列。

首先給一個簡單的想法，如果知道任意兩點是否可以交換（包括間接交換），那麼，我們只需要從最小的數開始一個一個的確定最小的位置 ，就可以得出答案了。如果要知道間接是否可交換，相當於求，任意兩點間，最短路是否可爲1，用floyed算法求出，複雜度爲o(n ^3)

#define N 305
#define M 100005
#define maxn 205
#define MOD 1000000000000000007
int n,b[N][N],p[N],a[N];
char str[N];
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;
}

第二種方法。用並查集優化，由於，在同一個集合內的點，明顯，只需要從小到大排（因爲同一集合，經過一點變換，一定可以生成各種排列），就一定是最優解了。複雜度，爲O(n * n);

#define N 305
#define M 100005
#define maxn 205
#define MOD 1000000000000000007
int n,p[N],num[N];
char str[N];
vector<int> g[N];
struct UnionFind{
    int fa[N];
    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;
}