拓扑序|Topological Order_1146

最新推荐文章于 2021-01-24 18:03:25 发布

原创最新推荐文章于 2021-01-24 18:03:25 发布 · 250 阅读

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本文深入探讨了拓扑排序算法的原理与应用，通过详细解释拓扑序的定义及其在有向无环图中的作用，提供了算法的具体实现步骤。文章还包含了一个基于2018年研究生入学考试问题的实例，旨在帮助读者理解如何通过编程测试给定的选项是否符合拓扑排序的要求。

描述

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.

指定输入

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.

指定输出

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.

输入样例

输出样例

3 4

分析

这道题我做得答案错了,但思路是正确的,程序还有bug,鉴于现在时间已经很晚了,先把过程记录下来,以后在做修改

分析点一:拓扑序的定义

拓扑序:若不存在回路即为有向无环图,所有节点可排成线性序列,使得节点的前驱节点都排在该节点的前面

分析点二:拓扑序算法

首先扫描每一个节点,如果入度等于0(即无前驱节点),则入队
当队列不为空时,出队,接下来处理与这个顶点相邻接的点
找到所有与当前节点相邻接的点,删除边,即入度减一,如过入度为0,加入到队列中

分析点三:c++实现的拓扑排序

/* 邻接表存储 - 拓扑排序算法 */
 
bool TopSort( LGraph Graph, Vertex TopOrder[] )
{ /* 对Graph进行拓扑排序,  TopOrder[]顺序存储排序后的顶点下标 */
    int Indegree[MaxVertexNum], cnt;
    Vertex V;
    PtrToAdjVNode W;
       Queue Q = CreateQueue( Graph->Nv );
  
    /* 初始化Indegree[] */
    for (V=0; V<Graph->Nv; V++)
        Indegree[V] = 0;
         
    /* 遍历图，得到Indegree[] */
    for (V=0; V<Graph->Nv; V++)
        for (W=Graph->G[V].FirstEdge; W; W=W->Next)
            Indegree[W->AdjV]++; /* 对有向边<V, W->AdjV>累计终点的入度 */
             
    /* 将所有入度为0的顶点入列 */
    for (V=0; V<Graph->Nv; V++)
        if ( Indegree[V]==0 )
            AddQ(Q, V);
             
    /* 下面进入拓扑排序 */ 
    cnt = 0; 
    while( !IsEmpty(Q) ){
        V = DeleteQ(Q); /* 弹出一个入度为0的顶点 */
        TopOrder[cnt++] = V; /* 将之存为结果序列的下一个元素 */
        /* 对V的每个邻接点W->AdjV */
        for ( W=Graph->G[V].FirstEdge; W; W=W->Next )
            if ( --Indegree[W->AdjV] == 0 )/* 若删除V使得W->AdjV入度为0 */
                AddQ(Q, W->AdjV); /* 则该顶点入列 */ 
    } /* while结束*/
     
    if ( cnt != Graph->Nv )
        return false; /* 说明图中有回路, 返回不成功标志 */ 
    else
        return true;
}

我的代码

v, e = map(int, input().split(" "))
v_set = set({})
v_map = dict({})
for i in range(e):
    a, b = map(int, input().split(" "))
    v_set.add(b)
    if v_map.get(a) is None:
        temp = list()
        temp.append(b)
        v_map[a] = temp
    else:
        v_map[a].append(b)

res = []
v_container = []

for j in range(1, v + 1):
    if j not in v_set:
        v_container.append(j)

if len(v_container) > 0:
    p = set(v_container)
    res.append(p)

while True:
    t = []
    while len(v_container) > 0:
        temp = v_container.pop()
        s = v_map.pop(temp)
        t.append(s)

    v_set = set({})
    v_container = []
    for value in v_map.values():
        for n in value:
            v_set.add(n)
    temp_set = set({})
    if len(v_set) != 0:
        for d in t:
            for e in d:
                if e not in v_set and len(v_set) != 0:
                    temp_set.add(e)
        v_container = list(temp_set)
    else:
        for d in t:
            for e in d:
                temp_set.add(e)
        res.append(temp_set)

    if len(v_container) == 0:
        break

    po = set(v_container)
    res.append(po)

h = int(input())
cnt = 0
result_set = set({})
for g in range(h):
    s = list(map(int, input().split(" ")))
    cnt += 1
    count = 0
    for k in res:
        for l in range(len(k)):
            if s[count + l] not in k:
                result_set.add(cnt - 1)
                break
        count = count + len(k)

result_set = list(result_set)
for z in range(len(result_set)):
    if z == len(result_set) - 1:
        print(result_set[z], end="")
    else:
        print(result_set[z], end=" ")

??正文结束??