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  • PAT (Advanced Level) Practice 1146 Topological Order (25分) (拓扑序列的判断)

    1.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.

    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

    2.题目分析

    1.拓扑序列的判断:拓扑序列,即有向无环,每次找一个入度为零的点,将所有和它相连的点的入度减一(删除相连的边),重复此步骤(有n个点,每次选择一个入度为0的点,进行n次)。如果这时还有入度不为零的点,证明有环。

    2.注意要是进行多次判断的话,记得不能动原来的in数组,要另外设置备份数组进行修改判断

    3.代码

    #include<iostream>
    #include<vector>
    #include <algorithm>
    using namespace std;
    #define MAX 1050
    int n, m,k;
    int edges[MAX][MAX];
    int in[MAX];
    int main()
    {
    	scanf("%d %d", &n, &m);
    	int a, b;
    	for (int i = 1; i <= m; i++)
    	{
    		scanf("%d %d", &a, &b);
    		edges[a][b] = 1;
    		in[b]++;
    	}
    	scanf("%d", &k);
    	int space = 0;
    	for (int i = 1; i <= k; i++)
    	{
            bool ok=true;
    		vector<int>temp(in, in + n + 1);
    		for (int j = 1; j <= n; j++)
    		{
    			int c;
    			scanf("%d", &c);
    			if (temp[c] != 0)
    			ok=false;
    			for (int s = 1; s <= n; s++)
    				if (edges[c][s] != 0)
                    temp[s]--;
    		}
                if(!ok)
                {
    				if (space == 0) { printf("%d", i - 1); space++; }
    				else printf(" %d", i - 1);
    			}
    	}
    }
    
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  • 原文地址:https://www.cnblogs.com/Jason66661010/p/12788888.html
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