我已经设置了一个代码,该代码将基于与节点之间的每个边缘相关联的值来导航图形。每条边都有一个与之相关的颜色和类型,如果颜色或类型与最后一种颜色或类型匹配,我应该只跟随BFS中的边。第一个颜色/类型由遵循的第一个边缘设置。但是,当我运行我的特定代码时,我在设置中的某个地方得到一个无法解决的infite循环。
我已经尝试设置不同的循环样式并使用迭代器而不是当前的for循环遍历我的列表,两者都不起作用,两者都会导致相同的错误。
队列Q;
Q.push(neededNode);
string lastType = "";
string lastColor = "";
while (!Q.empty()){
node u = Q.front();
Q.pop();
for(auto& itr : adjacencyList[u.city]){ //cycle through the adjacency list
for (auto& entry: nodeList){ //find the node for the entry
if (entry.city == (itr).city){
//Initial condition for setting color/type to follow
if(lastType == "" || lastColor == ""){
lastColor = (itr).color;
lastType = (itr).type;
entry.state = true;
entry.distance = u.distance +1;
entry.parent = u.city;
cout << u.city << " " << lastColor << " " << lastType << endl;
//If Types match
}else if(lastType == (itr).type){
lastColor = (itr).color;
lastType = (itr).type;
entry.state = true;
entry.distance = u.distance +1;
entry.parent = u.city;
cout << u.city << " " << lastColor << " " << lastType << endl;
//If colors match
}else if(lastColor == (itr).color){
lastColor = (itr).color;
lastType = (itr).type;
entry.state = true;
entry.distance = u.distance +1;
entry.parent = u.city;
cout << u.city << " " << lastColor << " " << lastType << endl;
}
Q.push(entry);
}
}
}
}
理想情况下,代码应该在完成时运行并停止运行,并且我应该能够将父值跟随到正确的路径。目前,我得到一个无限循环。
标记您已经看过的节点,确保您不再访问它们(例如使用.state):
if (entry.city == (itr).city && !entry.state) {
//Initial condition for setting color/type to follow