1146 Topological Order（25 分）

最新推荐文章于 2020-09-05 10:48:03 发布

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最新推荐文章于 2020-09-05 10:48:03 发布

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分类专栏：数据结构-----拓扑排序

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本文介绍了一个用于验证给定序列是否为有效拓扑排序的程序设计问题。通过输入一个有向图及其若干顶点排列，程序将判断哪些排列不符合拓扑排序的规则。

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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

这题题虽然打着扩扑排序的名号，但是其实用扩扑排序并没有用到，但是用到了扩扑排序的思想，对于要验证的例子，我们一个一个读入，来看看它的入度是否为0，如果不为0，那么它就不是扩扑排序

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<map>
using namespace std;
const int N = 1e3 + 10;
vector<int>v[N];
vector<int>s;
int indeg[N];
int main()
{
	int n, m;
	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);
		indeg[b]++;
	}
	int k;
	scanf("%d", &k);
	for (int i = 0;i < k;i++) {
		vector<int>p(indeg, indeg + n + 1);
		int flag = 1;
		for (int j = 0;j < n;j++) {
			int x;
			scanf("%d", &x);
			if (p[x] != 0) flag = 0;
			for (int z = 0;z < v[x].size();z++) {
				p[v[x][z]]--;
			}
		}
		if (flag == 0) {
			s.push_back(i);
		}
	}
	int flag = 0;
	for (int i = 0;i < s.size();i++) {
		if (flag) printf(" ");
		printf("%d", s[i]);
		flag = 1;
	}
	printf("\n");
	return 0;
}