1146 Topological Order
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
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
输出样例
3 4
分析
这道题我做得答案错了,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=" ")