图论——拓扑排序

Genealogical tree

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3150		Accepted: 2130		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

5
0
4 5 1 0
1 0
5 3 0
3 0

Sample Output

2 4 5 3 1

POJ上很基础一道拓扑排序题，这种题一般有两种做法：

第一种：

（1）从AOV网中选择一个没有前驱的顶点并且输出它；

（2）从AOV网中删除该顶点，并且上去所有该顶点为尾的弧；

（3）重复上述两步，直到全部顶点都被输出，或者AOV网中不存在没有前驱的顶点。

第二种：

     使用深度优先搜索（DFS），并标记每一个节点的第一次访问（pre）和最后一次访问时间（post），最后post的逆序就是DAG的拓扑排序，其实也是节点在进行DFS搜索时，出栈的逆序就是拓扑排序。

这里我用了第一种，当然，追求效率的话可以套用模板，下面给的是第一种方法的模板

/*==================================================*\
 | 拓扑排序
 | INIT:edge[][] 置为图的邻接矩阵;count[0…i…n-1]: 顶点i的入度.
\*==================================================*/
void TopoOrder(int n){
 int i, top = -1;
 for( i=0; i < n; ++i )
 if( count[i] == 0 ){ // 下标模拟堆栈
    count[i] = top; top = i;
 }
 for( i=0; i < n; ++i )
 if( top == -1 ) { printf(" 存在回路\n"); return ; }
 else{
 int j = top; top = count[top];
 printf("%d", j);
 for( int k=0; k < n; ++k )
 if( edge[j][k] && (--count[k]) == 0 ){
 count[k] = top; top = k;
 }
 }
}

直接上代码：

#include "iostream"

#include "cstring"
#include "algorithm"
#include "stdio.h"
using namespace std;

void TopoOrder(int n);
int n,edge[101][101],counts[101],answer[101];
int main() {

    memset(edge,0,sizeof(edge));
    cin>>n;
    int j,i = 1;
    for(; i<=n; i++) {
        int k;
        j = 0 ;
        while(cin>>k && k!=0) {

            edge[k][i] = 1;
            j++;
        }
        counts[i] = j;
    }


    TopoOrder(n);
}
void TopoOrder(int n) {
    int i, top = -1;
    for( i=1; i <= n; ++i )
        if( counts[i] == 0 ) { // 下标模拟堆栈
            counts[i] = top;
            top = i;
        }
    for( i=1; i <= n; ++i )
        if( top == -1 ) {
            printf(" 存在回路\n");
            return ;
        } else {
            int j = top;
            top = counts[top];
            //printf("%d", j);
            answer[i] = j;
            for( int k=1; k <= n; ++k )
                if( edge[j][k] && (--counts[k]) == 0 ) {
                    counts[k] = top;
                    top = k;
                }
        }
    for(int times=n; times>0; --times) {
        cout<<answer[times];
        if(n != 1)
            cout<<" ";
    }
}