[CareerCup] 8.6 Jigsaw Puzzle 拼图游戏

8.6 Implement a jigsaw puzzle. Design the data structures and explain an algorithm to solve the puzzle. You can assume that you have a f itsWith method which, when passed two puzzle pieces, returns true if the two pieces belong together.

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alt="" width="176" height="132" />

这道题让我们设计一个拼图游戏，根据书上的解释，如上图所示是一种最基本的拼图游戏，每一片有四条边，总共有三种边，inner, outer, 和 flat的，角落的一片有两个flat的边，中间的片没有flat的边。那么我们需要一个边类Edge，还需要一个片类Piece，和一个拼图类Puzzle，在拼图类里用sort来初始化参数，再用solve来完成拼图。这里不得不吐槽一下，这道题在网上下载的代码里面都木有，而且书上的代码有很多小错误，改正后参见代码如下：

class Edge;

class Piece {
public:
    vector<Edge*> _edges;
    bool isCorner() {} // ...
};

enum Type { inner, outer, flat };

class Edge {
public:
    Piece *_parent;
    Type _type;
    int _index;
    Edge *_attachedTo;
    bool fitsWith(Edge *edge) {} // ...
};

class Puzzle {
public:
    vector<Piece*> _pieces;
    vector<vector<Piece*> > _solution;
    vector<Edge*> _inners, _outers, _flats;
    vector<Piece*> _corners;
    void sort() {
        for (Piece *p : _pieces ) {
            if (p->isCorner()) _corners.push_back(p);
            for (Edge *e : p->_edges) {
                if (e->_type == Type::inner) _inners.push_back(e);
                if (e->_type == Type::outer) _outers.push_back(e);
            }
        }
    }
    void solve() {
        Edge *currentEdge = getExposedEdge(_corners[]);
        while (currentEdge != nullptr) {
            vector<Edge*> opposites = currentEdge->_type == Type::inner ? _outers : _inners;
            for (Edge *e : opposites) {
                if (currentEdge->fitsWith(e)) {
                    attachEdges(currentEdge, e);
                    removeFromList(currentEdge);
                    removeFromList(e);
                    currentEdge = nextExposedEdge(e);
                    break;
                }
            }
        }
    }
    void removeFromList(Edge *edge) {
        if (edge->_type == Type::flat) return;
        vector<Edge*> array = edge->_type == Type::inner ? _inners : _outers;
        for (vector<Edge*>::iterator it = array.begin(); it != array.end(); ++it) {
            if (*it == edge) {
                array.erase(it);
                break;
            }
        }
    }
    Edge* nextExposedEdge(Edge *edge) {
        int next_idx = (edge->_index + ) % ; // Opposite edge
        Edge *next_edge = edge->_parent->_edges[next_idx];
        if (isExposed(next_edge)) {
            return next_edge;
        }
        return getExposedEdge(edge->_parent);
    }
    void attachEdges(Edge *e1, Edge *e2) {
        e1->_attachedTo = e2;
        e2->_attachedTo = e1;
    }
    bool isExposed(Edge *edge) {
        return edge->_type != Type::flat && edge->_attachedTo == nullptr;
    }
    Edge* getExposedEdge(Piece *p) {
        for (Edge *e : p->_edges) {
            if (isExposed(e)) return e;
        }
        return nullptr;
    }
};