【POJ 2367】 Genealogical tree 拓扑排序_d - genealogical tree poj - 2367 (拓扑排序) java视频-优快云博客

该博客介绍了火星生物混乱的血缘关系系统，并提出了一个问题：如何确保在行星议会上，每个成员发言都在其所有后代之前。问题转化为拓扑排序，给定一个表示亲属关系的图，寻找一个排序序列满足条件。输入包含一个数字n表示成员数量，接下来n行列出每个成员的子女。博主提供了一个拓扑排序的模板作为解答。

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Genealogical tree

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8455		Accepted: 5437		Special Judge

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

Sample Output

2 4 5 3 1

题意：先输入一个数字n，代表有n个顶点，接下来n行，第i行代表第i个顶点和其他顶点相连的点，以0结束。

题解：拓扑排序模板题

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int maxn=500+7;
int graph[maxn][maxn];          //图的邻接矩阵
int digree[maxn];  //存储每个顶点的入度
int n,v;
int main()
{
    while(scanf("%d",&n)==1){
        for(int i=1;i<=n;i++){
            while(scanf("%d",&v)==1&&v){
                    if(!graph[i][v]){
                        graph[i][v]=1;
                        digree[v]++;       //入度+1
                    }
            }
        }
        queue<int>q;
        for(int i=1;i<=n;i++){         //将入度为0的点 加入队列
            if(digree[i]==0) q.push(i);
        }
        int first=1;
        while(!q.empty()){            //利用BFS 完成 拓扑排序
            int cur=q.front();
            q.pop();
            if(first==1){
                printf("%d",cur);
                first=0;
            }
            else printf(" %d",cur);
            for(int i=1;i<=n;i++){
                if(graph[cur][i]){
                    digree[i]--;
                    if(digree[i]==0)  q.push(i);   //入度为0 加入队列
                }
            }
        }
        printf("\n");
    }
    return 0;
}