Question

我目前正在使用二叉搜索树处理电影数据库项目。树中的每个节点都包含电影的标题，电影制作的年份以及包含所有演员姓名的链接列表。这些信息都是从.txt文件中读取的。

我已经创建了一些功能，但是我遇到了麻烦。我正在尝试创建一个函数，使用户能够输入一个字符串，该字符串将是actor的名字和姓氏，然后遍历树节点。如果在节点内的链接列表中找到了actor名称，则该函数将打印出该影片的标题。

我以前从未使用过STL，所以我不知道访问元素是否与手动创建列表相同，这就是我遇到问题的原因。

任何帮助将不胜感激。到目前为止，这是我的代码：

#include <iostream>
#include <cstdlib>
#include <fstream>
#include <string>
#include <fstream>
#include <list>

using namespace std;

typedef struct treeNode{
    int year;
    string title;
    list<string> actors;
    treeNode* left;
    treeNode* right;

}* treePtr;

treePtr root = NULL;

treePtr fillTree(treePtr p, int y, string t, list<string> a);
void print_titles(treePtr root);
void print_before_year(treePtr root, int key);
void print_actors_movies(???);


int main(){

    ifstream inFile ("movies.txt");
    string x;
    treePtr root = NULL;
    int count = 0;
    if(inFile.is_open()){

        while(getline (inFile, x)){
            if(x == ""){
                continue;
            }
            count++;

            int index = x.find_last_of(" ");
            string title = x.substr(0, index);
            string year = x.substr(index + 2, x.length() - index - 3);

            list<string> actor;
            int counter = 0;

            while(getline(inFile, x)){
                if(x == ""){
                    break;
                }
                else{
                    actor.push_back(x);
                }
            }

            root = fillTree(root, atoi(year.c_str()), title, actor);
        }
    }
    int choice;

    do{

        cout <<"\nWelcome to the Movie Store. Would you like to: \n(1)
        See what movies are available? \n(2) Search for an actor? \n(3)
            Search for a year?   \n(4) Search for a movie? \n(0) Exit the Store" << endl;

                cin >> choice;

        switch(choice){
            case 0:
                cout << "Thank you come again." << endl;
                break;
            case 1:
                print_titles(root);
                break;
            case 2:

            case 3:
                int year;
                cout << "Please enter the year you wish to search for: " << endl;
                cin >> year;
                cout << endl;
                cout << "Films made before " << year << ":" << endl;
                print_before_year(root, year);
                break;
            case 4:

            default:
                cout << "Try again." << endl;
        }

    } while(choice != 0);

    return 0;
}


treePtr fillTree(treePtr p, int y, string t, list<string> a){
    treePtr n = new treeNode;
    n->year = y;
    n->title = t;
    n->actors = a;
    n->left = NULL;
    n->right = NULL;
    if(p == NULL){
        p = n;
    }
    else{
        treePtr prev = p;
        treePtr curr = p;
        while(curr != NULL){
            if(curr->year > y){
                prev = curr;
                curr = curr->left;
            }
            else{
                prev = curr;
                curr = curr->right;
            }
        }
        if(prev->year > y){
            prev->left = n;
        }
        else{
            prev->right = n;
        }


    }

    return p;
}

void print_titles(treePtr root){
    if(root == NULL) return;
    if(root->left) print_titles(root->left);
    cout<<root->title<<endl;
    if(root->right) print_titles(root->right);
}

void print_before_year(treePtr root, int key){

    if(root == NULL) return;
    if(root->left) print_before_year(root->left, key);
    if(root->year < key){
        cout << root->title << endl;
    }
    else return;
    if(root->right) print_before_year(root->right, key);
}

void print_actors_movies(???){

}

以下是.txt文件以防万一：http://www2.cs.uidaho.edu/~bruceb/cs121/Assignments/movies.txt

Answer 1

如果您有treeNode，并且想要查找名称actor，则可以使用

bool hasActor(treeNode const* node, std::string const& name) {
    return node->actors.end()
        != std::find(node->actors.begin(), node->actors.end(), name);
}

函数std::find()使用{{1}将begin和end范围内每个位置标识的值与作为最后一个参数传递的value进行比较}。如果此表达式返回*it == value，则返回true。如果此表达式没有返回it的位置，则返回true。

Answer 2

简单的广度优先搜索和std::find()怎么样？

bool printTitleForActor(Node *root, const std::string &actor)
{
    if (!root)
        return false;

    if (std::find(root->actors.begin(), root->actors.end(), actor) != root->actors.end()) {
        std::cout << root->title << std::endl;
        return true; // only if you want to terminate upon first find
    }

    return printTitleForActor(root->left) || printTitleForActor(root->right);
}

需要帮助迭代STL列表

2 个答案: