问题描述:
1146. Topological Order (25)
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.
![](http://nos.patest.cn/qp_p5m3xdiy3a6.jpg)
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 6Sample Output:
3 4
只要按照拓扑排序的算法一步一步地验证就行;即设置一个出度和入度数组,每次在遇到顶点时,检查此顶点的入度是否为0,是则将所有此顶点指向的顶点入度减1,继续检查下一个,否则,输出序号。
AC代码:
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#include<bits/stdc++.h> using namespace std; vector<vector<int> > v; int main() { ios::sync_with_stdio(false); // freopen("data.txt","r",stdin); int n,m,k,c1,c2,x; cin>>n>>m; v.resize(n); vector<int> vi(n,0); for(;m--;) { cin>>c1>>c2; c1--; c2--; v[c1].emplace_back(c2); vi[c2]++; } cin>>m; bool fp=true; for(int i=0;i<m;i++) { vector<int> vii=vi; int x,nn; bool flag=true; for(nn=n;nn--;) { cin>>x; x--; if(flag) { if(vii[x]) { flag=false; } else for(int i=0;i<v[x].size();i++) vii[v[x][i]]--; } } if(!flag) if(fp) { cout<<i; fp=0; } else cout<<" "<<i; } return 0; } |