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题目

This is a problem given in the Graduate Entrance Exam in 2018: Which of the following is NOT a topological order obtained from the given directed graph? Now you are supposed to write a program to test each of the options.

Input Specification:
Each input file contains one test case. For each case, the first line gives two positive integers N (≤ 1,000), the number of vertices in the graph, and M (≤ 10,000), the number of directed edges. Then M lines follow, each gives the start and the end vertices of an edge. The vertices are numbered from 1 to N. After the graph, there is another positive integer K (≤ 100). Then K lines of query follow, each gives a permutation of all the vertices. All the numbers in a line are separated by a space.
Output Specification:
Print in a line all the indices of queries which correspond to "NOT a topological order". The indices start from zero. All the numbers are separated by a space, and there must no extra space at the beginning or the end of the line. It is graranteed that there is at least one answer.
Sample Input:
6 8
1 2
1 3
5 2
5 4
2 3
2 6
3 4
6 4
5
1 5 2 3 6 4
5 1 2 6 3 4
5 1 2 3 6 4
5 2 1 6 3 4
1 2 3 4 5 6
Sample Output:
3 4

题目分析

已知有向无环图，顶点和边，判断一系列序列是否是拓扑序列
拓扑序列：

• 每次从图中取一个入度为0的顶点，并将该顶点指向的顶点的入度减1
• 继续取出入度为0的顶点，循环一直到图中所有顶点被取出

解题思路

1. 适合用邻接表存储有向图（访问边遍历复杂度低），并记录每个顶点的入度
2. 判断是否为拓扑排序
   2.1 依次输入拓扑序列的顶点，判断入度是否为0
   若入度不为0，isTop=false，即不是拓扑序列
   若入度为0，将当前节点指向的顶点的入度-1
   2.2 依次循环执行，直到序列中所有顶点被校验

易错点

1. 在解题思路2.1中判断入度不为0，不能直接break，因为要将序列接收完
2. 每次新输入一个序列时，需要重新拷贝一份图各个顶点的入度信息，重新计算

Code

#include <iostream>
#include <vector>
using namespace std;
int main(int argc,char * argv[]) {
    int n,m,a,b,k,v,flag=0;
    scanf("%d %d",&n,&m);
    vector<int> edge[n+1];
    int indegree[n+1]= {0};
    for(int i=0; i<m; i++) {
        scanf("%d %d",&a,&b);
        edge[a].push_back(b);
        indegree[b]++;
    }
    scanf("%d",&k);
    for(int i=0; i<k; i++) {
        int isTop = 1;
        vector<int> cindegree(indegree, indegree+n+1);
        for(int j=1; j<=n; j++) {
            scanf("%d",&v);
            if(cindegree[v]!=0) isTop=0;
            for(int r=0; r<edge[v].size(); r++) {
                --cindegree[edge[v][r]];
            }
        }
        if(isTop==0) {
            printf("%s%d",flag==1?" ":"",i);
            flag=1;
        }
    }
    return 0;
}

PAT Advanced 1146 Topological Order (25) [拓扑排序]

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原文地址：https://www.cnblogs.com/houzm/p/12391372.html