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.
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.
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.
题意：先输入顶点数和边数，再输入边的情况，再输入查询次数，和每次查询所输入的拓扑序列 思路：拓扑排序可能会很容易想到栈，但是看了柳神的博客以后才发现用一个二维vector邻接矩阵和一个一维的in数组表示每个顶点的入度就能搞定。这里的几个坑点是最好设置新的vector tin(in,in+n+1)这里一定要加一，否则就出错了，还有一个坑点是每个数之间要空格输出，应该要将控制空格的标识flag放在for循环的外层，放到内存一直报格式错误。一个比较巧妙的点是只要将n个顶点遍历过去，判断他的入度是不是0，引入judge标识，不等于0就用增强for循环把与这个点相邻的点的入度都减一，最后只需要看judge是不是为一就可以了，为一就说明是拓扑序列,continue，不唯一就打印出来。
#include<iostream> #include<cstdio> #include<stack> #include<queue> #include<vector> #include<algorithm> #include<cstring> #include<string> #include<cmath> #include<set> #include<map> #include<cctype> #include<cstdlib> #include<ctime> #include<unordered_map> using namespace std; int main() { int n,m,c; vector<int> v[1010]; //这个flag=0要放在外面 int flag = 0; int in[1010]; scanf("%d%d",&n,&m); for(int i = 0;i < m;i++) { int a,b; scanf("%d%d",&a,&b); v[a].push_back(b); in[b]++; } int k; scanf("%d",&k); for(int i = 0;i < k;i++) { int judge = 1; //vector最好设置大一点 vector<int> tin(in,in + n + 1); for(int i = 0;i < n;i++) { scanf("%d",&c); if(tin[c] != 0) judge = 0; for(int it : v[c]) tin[it]--; } if(judge == 0) { printf("%s%d", flag == 1 ? " ": "", i); flag = 1; } else if (judge == 1) { continue; } } return 0; }