POJ - 2367(拓扑排序)

最新推荐文章于 2019-09-01 16:25:04 发布

20172674

最新推荐文章于 2019-09-01 16:25:04 发布

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分类专栏：数据结构

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该博客介绍了如何解决一个关于火星人血缘关系系统的拓扑排序问题。在火星议会中，为了维持秩序，需要确保每个成员发言在他们的子孙之前。输入包含议会成员及其子女信息，任务是确定一个满足条件的发言顺序。博主解释了问题的本质是辈分从大到小的拓扑排序，并提供了实现该排序的AC代码。

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题目链接：http://poj.org/problem?id=2367

题目：

Description

The system of Martians' blood relations is confusing enough. Actually, Martians bud when they want and where they want. They gather together in different groups, so that a Martian can have one parent as well as ten. Nobody will be surprised by a hundred of children. Martians have got used to this and their style of life seems to them natural.
And in the Planetary Council the confusing genealogical system leads to some embarrassment. There meet the worthiest of Martians, and therefore in order to offend nobody in all of the discussions it is used first to give the floor to the old Martians, than to the younger ones and only than to the most young childless assessors. However, the maintenance of this order really is not a trivial task. Not always Martian knows all of his parents (and there's nothing to tell about his grandparents!). But if by a mistake first speak a grandson and only than his young appearing great-grandfather, this is a real scandal.
Your task is to write a program, which would define once and for all, an order that would guarantee that every member of the Council takes the floor earlier than each of his descendants.

Input

The first line of the standard input contains an only number N, 1 <= N <= 100 — a number of members of the Martian Planetary Council. According to the centuries-old tradition members of the Council are enumerated with the natural numbers from 1 up to N. Further, there are exactly N lines, moreover, the I-th line contains a list of I-th member's children. The list of children is a sequence of serial numbers of children in a arbitrary order separated by spaces. The list of children may be empty. The list (even if it is empty) ends with 0.

Output

The standard output should contain in its only line a sequence of speakers' numbers, separated by spaces. If several sequences satisfy the conditions of the problem, you are to write to the standard output any of them. At least one such sequence always exists.

Sample Input
5
0
4 5 1 0
1 0
5 3 0
3 0
Sample Output
2 4 5 3 1

题目意思：

题目给出辈分关系要求从大到小输出辈分

拓扑排序每次找出入度为0的点然后输出并删除它的所有出度，再找新的入度为0的点

AC代码：

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;
const int maxn = 0X3f3f3f3f ;

int main()
{
    int n,child[105][105],parent[105];
    scanf("%d",&n);
    memset(child,0,sizeof(child));
    memset(parent,0,sizeof(parent));
    for(int i=1 ; i<=n ; i++)
    {
        for(int j=1 ; ; j++)
        {
            scanf("%d",&child[i][j]);
            if(child[i][j] == 0) break;
            parent[child[i][j]]++;
        }
    }
    for(int i=1 ; i<=n ; i++)
    {
        for(int j=1 ; j<=n ; j++)
        {
            if(parent[j] == 0)
            {
                if(i == 1) printf("%d",j);
                else printf(" %d",j);
                parent[j] = maxn;
                for(int k=1;;k++)
                {
                    if(child[j][k] == 0)break;
                    parent[child[j][k]]--;
                }
                break;
            }
        }
    }

}