zoukankan      html  css  js  c++  java
  • New Year Permutation(Floyd+并查集)

    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

    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


    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

    这题也是对并查集数据结构的一种很好的理解,不过忽略了并查集分组的性质,而只关注了并查集的连通性
    即通过邻接矩阵得到扩展边与边的连通关系

    #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
    ",p[n-1]);
        }
        return 0;
    }
  • 相关阅读:
    C语言 · 最小公倍数
    SSH实战 · SSH项目开发环境搭建
    C语言 · 回文数
    C语言 · 特殊回文数
    C语言 · 查找整数
    SSH实战 · SSH项目中怎么玩验证码
    SSH实战 · JAVA发送邮件相关
    SSH实战 · AJAX异步校验
    C语言 · 打印1-200之间的素数
    Jenkins权限设计错误解决办法
  • 原文地址:https://www.cnblogs.com/immortal-worm/p/4975907.html
Copyright © 2011-2022 走看看