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c++ - C++ 中的迭代加深搜索

转载 作者:太空宇宙 更新时间:2023-11-04 13:52:03 25 4
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好的,所以,首先,我并不知道我在用迭代加深做什么。我一直在努力让这段代码工作,但我做不到。我在网上看了看,找不到任何关于此搜索的 C++ 引用资料。

void Graph::IDS(int x, int required, int depth = 1)
{
if(x == required) return;

cout << "Iterated Deepening Search for " << required << ", starting from vertex " << x << " : " << endl;

IDS_util(x, required, depth);

cout << endl;
}

void Graph::IDS_util(int x, int required, int depth)
{
stack s;
bool *visited = new bool[n+1];
int i, j, k;

for(i = 0; i <= n; i++)
visited[i] = false;

cout << "Depth = " << depth << ": ";

visited[x] = true;

for (int c = 1; c <= n; c++){
s.push(x);

if(isConnected(x, c) && !visited[c])
{
for (j = 0; j < depth; j++){
k = s.pop();

if(k == required) return;

cout << "[" << k <<"] ";

for (i = n; i >= 0 ; --i)
if (isConnected(k, i) && !visited[i]) {
s.push(i);
visited[i] = true;
}
}
}
}

if(depth == n) return;

cout << endl;

IDS_util(x, required, depth+1);
}

邻接矩阵的输出:

0,1,1,1,0,0,0,0,0
0,0,0,0,1,0,0,0,0
0,0,0,0,0,1,1,0,0
0,0,0,0,0,0,0,1,0
0,0,0,0,0,0,0,0,1
0,0,0,0,0,0,0,0,0
0,0,0,0,0,0,0,0,0
0,0,0,0,0,0,0,0,0
0,0,0,0,0,0,0,0,0

这是此图的定向版本:

         [1] 
/ | \
[2] [3] [4]
/ / \ \
[5] [6] [7] [8]
/
[9]

是:

Iterated Deepening Search for 7, starting from vertex 1 : 
Depth = 0:
Depth = 1: [1]
Depth = 2: [1] [2]
Depth = 3: [1] [2] [5]
Depth = 4: [1] [2] [5] [9]
Depth = 5: [1] [2] [5] [9] [3]
Depth = 6: [1] [2] [5] [9] [3] [6]
Depth = 7: [1] [2] [5] [9] [3] [6]

理论上我知道搜索应该做什么,我可以稍微说出我的搜索在做什么,但我不知道如何修复它。任何人都可以提供任何帮助,我们将不胜感激。

最佳答案

我几乎可以肯定这不是最有效的方法,但我找到了一种可行的方法。

void Graph::IDS(int x, int required)
{
if(x == required) return;

cout << "Iterated Deepening Search for " << required << ", starting from vertex " << x << " : " << endl;

for (int d = 0 ; d <= n ; d++)
if (IDS_util(x, required, d))
return;

cout << required << " was unable to be located via " << x << endl;
}



bool Graph::IDS_util(int x, int required, int depth){

if(x == required) return true;

stack s, x_child;
bool *visited = new bool[n+1];
int i,k, d, sub_k;

for(i = 0; i <= n; i++) visited[i] = false;

visited[x] = true;

for (i = n; i >= 0 ; --i)
if (isConnected(x, i))
x_child.push(i);

cout << '[' << x << "] ";

while(!x_child.isEmpty()){
k = x_child.pop();
s.push(k);

for(d = 0; d < depth; d++){
sub_k = s.pop();
if(sub_k == required) return true;

cout << '[' << sub_k << "] ";

for (i = 0; i <= n; i++){
if (isConnected(sub_k, i) && !visited[i]) {
if (i == required){
cout << "\n\n" << required << " is a child of " << sub_k << endl;
return true;
}

s.push(i);
visited[i] = true;
}
}
}
}
cout<<endl;
delete [] visited;

return false;
}

关于c++ - C++ 中的迭代加深搜索,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/23098396/

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