POJ 2367Genealogical tree（拓扑排序）

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

题意：给出n下面有五个节点的信息，每个节点的信息以0结尾。
看数据：
第１个节点没有儿子节点。
第２个节点有４,５,１三个儿子节点。
第３个节点有１　一个儿子节点。
第４个节点有５，３两个儿子节点。
第５个节点有３　一个儿子节点。

图就不画了。

然后将这些节点拓扑排序。

解题思路：
拓扑排序：
１.从ＡＯＶ网络中选择一个入度为０（即没有直接前驱）的顶点并输出。（入队列）
２．从ＡＯＶ网络中删除该顶点及该顶点发出的所有边。（与上一顶点连接的顶点的入度－１）
３．重复１.２直到找不到入度为０的顶点。

ＡＣ代码：

#include<stdio.h>#include<string.h>#include<queue>#include<algorithm>using namespace std;int mp[110][110];int in[110];int n;void toposort(){    int book[110]={0};    queue<int> Q;    for(int i=1;i<=n;i++)    {        if(in[i]==0) {book[i]=1;Q.push(i);}    }    int k=1;    while(!Q.empty())    {        int u=Q.front();        Q.pop();        printf("%d",u);        if(k++<n) printf(" ");        else printf("\n");        for(int i=1;i<=n;i++)            if(mp[u][i]==1) in[i]--;        for(int i=1;i<=n;i++)            if(!in[i]&&!book[i])            {                book[i]=1;                Q.push(i);                break;            }    }}int main(){    while(~scanf("%d",&n)&&n)    {        memset(mp,0,sizeof(mp));        memset(in,0,sizeof(in));        int u,v;        for(u=1;u<=n;u++)        {            while(~scanf("%d",&v)&&v)            {                mp[u][v]=1;                in[v]++;            }        }        toposort();    }    return 0;}