New Year Permutation（并查集+动态容器）

B. New Year Permutation 

time limit per test2 seconds 

memory limit per test256 megabytes 

inputstandard input 

outputstandard 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) 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 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.

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

题目意思：给出一个含有 n 个数的排列：p1, p2, ..., pn-1, pn。紧接着是一个 n * n 的矩阵A，当且仅当 Aij = 1 时，pi 与 pj 可以交换数值。现在问如何交换数值，使得最后得到的排列字典序最小。

　　比赛的时候不会做，看了Tutorial 1 的解法，觉得别人做得太巧妙了，出题者也出得很好 ^_^

　　可以看这个：http://codeforces.com/blog/entry/15488

    它是利用了并查集来做的。如果遇到 Aij = 1 的话，就将点 i 和 点 j 连一条边。最终会得到一些互不相交的集合。以第一组 test 来说吧～～～

　　

　　

　　这就表示位置 1、4、7 的数是可以交换的，位置 2、5 要保持原封不动，位置 3、6 的数可以交换。由于每个集合都有一个祖先，即第一层的那个数，依次为 7、2、5、6，然后就把每个集合的数放到以该集合祖先为首的 vector 数组里面，并把每个集合内的数从小到大排序。最后遍历位置 1 ～ n，根据每个位置的数所属的集合（以哪个祖先为首），依次输出。

#include<cstdio>#include<cstring>#include<algorithm>#include<vector>using namespace std;int a[305];int f[305];int c[305];vector<int>vec[305]; int find(int x){int i,j,r;r=x; while(f[r]!=r)      r=f[r];return r;}int root(int x,int y){int px=find(x);int py=find(y);if(f[px]!=f[py])   f[px]=py;}int main(){int n;scanf("%d",&n);memset(c,0,sizeof(c));   for(int i=1;i<=n;i++){scanf("%d",&a[i]);f[i]=i;    }for(int i=1;i<=n;i++){char s[305];scanf("%s",s+1);for(int j=i;j<=n;j++){if(s[j]=='1'){root(i,j);}}}    for(int i=1;i<=n;i++)    {    int v=find(i);    vec[v].push_back(a[i]);}for(int i=1;i<=n;i++){sort(vec[i].begin(),vec[i].end());  //排序 }for(int i=1;i<n;i++)printf("%d ",vec[find(i)][c[find(i)]++]);  //输出每个容器的数，输出的前先令c数组都为0，这样下次输入可按当前顺序 printf("%d\n",vec[find(n)][c[find(n)]++]);return 0;}