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 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 Input
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
思路：
用floyd先将左右潜在的可以连通的边在邻接矩阵中标记出来，然后像冒泡一样的给他们排序一下就可以了
要注意两点：
一是刚开始开数组的时候，要从0开始开，不然用scanf输入会很麻烦
二是floyd的三层for循环，注意k的顺序要在最前面
#include <iostream> #include <cstdio> #include <cstring> #include <queue> using namespace std; int n; int p[307]; char f[307][307]; int main() { while(~scanf("%d",&n)) { for(int i = 0;i < n;i++) scanf("%d",&p[i]); for(int i = 0;i < n;i++) { scanf("%s",f[i]); for(int j = 0;j < n;j++) if(f[i][j] == ‘1‘) f[i][j] = ‘1‘; } for(int k = 0;k < n;k++) for(int i = 0;i < n;i++) for(int j = 0;j < n;j++) if(f[i][k]==‘1‘ && f[k][j]==‘1‘) f[i][j] = ‘1‘; for(int i = 0;i < n;i++) for(int j = i+1;j < n;j++) if(f[i][j]==‘1‘ && p[j] < p[i]) swap(p[i],p[j]); for(int i = 0;i < n-1;i++) printf("%d ",p[i]); printf("%d\n",p[n-1]); } return 0; }
原文：http://www.cnblogs.com/immortal-worm/p/4975907.html