1146 Topological Order (25 分) 拓扑序列是否合法

Dear 丶 2021-10-30 08:16 210阅读 0赞

1146 Topological Order (25 分)

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.

![gre.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 6

### Sample Output: ###

3 4

**题意：**给定n个结点，m条有向边的拓扑图。

k个询问，问该序列是否合法。

输出不合法的序列编号（从0开始编号）

**思路：**用vector存有向边，用degree数组记录结点的入度。

每次询问重置td数组（结点的入度），每输入一个数就判断该结点的入度是否为0.

若否则该序列不合法，flag修改为1。

且将改点指向的结点的入度--

#include <stdio.h>
    #include <string.h>
    #include <string>
    #include <map>
    #include <vector>
    #include <queue>
    #include <math.h>
    #include <algorithm>
    #include <iostream>
    #define INF 0x3f3f3f3f
    using namespace std;
    vector<int> a[1005];//单向图 
    int degree[1005],td[1005];//入度
    vector<int> p;//输出的编号 
    int main()
    {
    	int n,m,i,j,l,k,x,y;
    	scanf("%d%d",&n,&m);
    	memset(degree,0,sizeof degree);
    	for(i=0;i<m;i++)
    	{
    		scanf("%d%d",&x,&y);
    		a[x].push_back(y);
    		degree[y]++;
    	}	
    	scanf("%d",&k);
    	for(i=0;i<k;i++)
    	{
    		for(j=1;j<=n;j++)
    		td[j]=degree[j];
    		int flag=0;
    		for(j=0;j<n;j++)
    		{
    			scanf("%d",&x);
    			if(td[x]!=0) flag=1;
    			for(l=0;l<a[x].size();l++)
    			td[a[x][l]]--;
    			
    		}
    		if(flag) p.push_back(i);
    	}
    	for(i=0;i<p.size();i++)
    	{
    		if(i)printf(" ");
    		printf("%d",p[i]);
    	}
    }

[gre.jpg]: https://imgconvert.csdnimg.cn/aHR0cHM6Ly9pbWFnZXMucHRhdXNlcmNvbnRlbnQuY29tLzVkMzVlZDJhLTRkMTktNGYxMy1iZjNmLTM1ZWQ1OWNlYmYwNS5qcGc