c++ - 两点之间网格中的最短路径。有一个陷阱

Question

我有这个问题，我必须通过仅向右或向下移动来找到从 A 点(总是左上角)到 B 点(总是右下角)的 NxM 网格中的最短路径。听起来很简单，是吗？那么问题来了:我只能移动我现在坐在的方 block 上显示的数字。让我举例说明:

2 5 1 2
9 2 5 3
3 3 1 1
4 8 2 7

在这个 4x4 网格中，最短路径需要 3 步，从左上角的 2 个节点向下走到 3，然后从右上角的 3 个节点走到 1，然后向下走 1 个节点到达目标。

[2] 5  1  2
 9  2  5  3
[3] 3  1 [1]
 4  8  2 [7]

如果不是最短路径，我也可以走这条路:

[2] 5 [1][2]
 9  2  5  3
 3  3  1 [1]
 4  8  2 [7]

不幸的是，这需要多达 4 个步骤，因此，我不感兴趣。那应该清楚一点。 现在关于输入。

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用户输入网格如下:

5 4      // height and width
2 5 2 2  //
2 2 7 3  // the
3 1 2 2  // grid
4 8 2 7  //
1 1 1 1  //

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作业

我已经仔细考虑过了，但没有比将输入的网格简化为未加权(或负权重)图并在其上运行类似 dijkstra 或 A*(或类似的东西)更好的解决方案了。嗯...这是我迷路的部分。我首先实现了一些东西(或者立即扔掉的东西)。它与 dijkstra 或 A* 或任何东西无关；只是简单的广度优先搜索。

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代码

#include <iostream>
#include <vector>

struct Point;

typedef std::vector<int> vector_1D;
typedef std::vector< std::vector<int> > vector_2D;
typedef std::vector<Point> vector_point;

struct Point {
    int y, x;
    vector_point Parents;
    Point(int yPos = 0, int xPos = 0) : y(yPos), x(xPos) { }

    void operator << (const Point& point) { this->Parents.push_back(point); }
};

struct grid_t {
    int height, width;
    vector_2D tiles;

    grid_t() // construct the grid
    { 
        std::cin >> height >> width; // input grid height & width

        tiles.resize(height, vector_1D(width, 0)); // initialize grid tiles

        for(int i = 0; i < height; i++)     //
            for(int j = 0; j < width; j++)  // input each tile one at a time
                std::cin >> tiles[i][j];    // by looping through the grid
    }
};

void go_find_it(grid_t &grid)
{
    vector_point openList, closedList;
    Point previous_node; // the point is initialized as (y = 0, x = 0) if not told otherwise
    openList.push_back(previous_node); // (0, 0) is the first point we want to consult, of course

    do
    {
        closedList.push_back(openList.back()); // the tile we are at is good and checked. mark it so.
        openList.pop_back(); // we don't need this guy no more

        int y = closedList.back().y; // now we'll actually
        int x = closedList.back().x; // move to the new point

        int jump = grid.tiles[y][x]; // 'jump' is the number shown on the tile we're standing on.

        if(y + jump < grid.height) // if we're not going out of bounds
        { 
            openList.push_back(Point(y+jump, x)); // 
            openList.back() << Point(y, x); // push in the point we're at right now, since it's the parent node
        }
        if(x + jump < grid.width) // if we're not going out of bounds
        { 
            openList.push_back(Point(y, x+jump)); // push in the new promising point
            openList.back() << Point(y, x); // push in the point we're at right now, since it's the parent node
        }
    }
    while(openList.size() > 0); // when there are no new tiles to check, break out and return
}

int main()
{
    grid_t grid; // initialize grid

    go_find_it(grid); // basically a brute-force get-it-all-algorithm

    return 0;
}

我可能还应该指出，运行时间不能超过 1 秒，并且网格的最大高度和宽度是 1000。所有的瓦片也是从 1 到 1000 的数字。

谢谢。

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编辑代码

#include <iostream>
#include <vector>

struct Point;

typedef std::vector<int> vector_1D;
typedef std::vector< std::vector<int> > vector_2D;
typedef std::vector<Point> vector_point;

struct Point {
    int y, x, depth;
    vector_point Parents;
    Point(int yPos = 0, int xPos = 0, int dDepth = 0) : y(yPos), x(xPos), depth(dDepth) { }

    void operator << (const Point& point) { this->Parents.push_back(point); }
};

struct grid_t {
    int height, width;
    vector_2D tiles;

    grid_t() // construct the grid
    { 
        std::cin >> height >> width; // input grid height & width

        tiles.resize(height, vector_1D(width, 0)); // initialize grid tiles

        for(int i = 0; i < height; i++)     //
            for(int j = 0; j < width; j++)  // input each tile one at a time
                std::cin >> tiles[i][j];    // by looping through the grid
    }
};

int go_find_it(grid_t &grid)
{
    vector_point openList, closedList;
    Point previous_node(0, 0, 0); // the point is initialized as (y = 0, x = 0, depth = 0) if not told otherwise
    openList.push_back(previous_node); // (0, 0) is the first point we want to consult, of course

    int min_path = 1000000;

    do
    {
        closedList.push_back(openList[0]); // the tile we are at is good and checked. mark it so.
        openList.erase(openList.begin()); // we don't need this guy no more

        int y = closedList.back().y; // now we'll actually move to the new point
        int x = closedList.back().x; //
        int depth = closedList.back().depth; // the new depth

        if(y == grid.height-1 && x == grid.width-1) return depth; // the first path is the shortest one. return it

        int jump = grid.tiles[y][x]; // 'jump' is the number shown on the tile we're standing on.

        if(y + jump < grid.height) // if we're not going out of bounds
        { 
            openList.push_back(Point(y+jump, x, depth+1)); // 
            openList.back() << Point(y, x); // push in the point we're at right now, since it's the parent node
        }
        if(x + jump < grid.width) // if we're not going out of bounds
        { 
            openList.push_back(Point(y, x+jump, depth+1)); // push in the new promising point
            openList.back() << Point(y, x); // push in the point we're at right now, since it's the parent node
        }
    }
    while(openList.size() > 0); // when there are no new tiles to check, break out and return false

    return 0;
}

int main()
{
    grid_t grid; // initialize grid

    int min_path = go_find_it(grid); // basically a brute-force get-it-all-algorithm

    std::cout << min_path << std::endl;
    //system("pause");
    return 0;
}

程序现在打印出正确答案。现在我必须优化(运行时间太大了)。关于这个有什么提示吗？优化是我最擅长的一件事。

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答案

最后，解决方案似乎只包含很少的代码。越少越好，我喜欢。感谢 Dejan Jovanović 提供的漂亮解决方案

#include <iostream>
#include <vector>
#include <algorithm>

struct grid_t {
    int height, width;
    std::vector< std::vector<int> > tiles;
    std::vector< std::vector<int> > distance;

    grid_t() // construct the grid
    { 
        std::cin >> height >> width; // input grid height & width

        tiles.resize(height, std::vector<int>(width, 0)); // initialize grid tiles
        distance.resize(height, std::vector<int>(width, 1000000)); // initialize grid tiles

        for(int i = 0; i < height; i++)     //
            for(int j = 0; j < width; j++)  // input each tile one at a time
                std::cin >> tiles[i][j];    // by looping through the grid
    }
};

int main()
{
    grid_t grid; // initialize grid

    grid.distance[0][0] = 0;
    for(int i = 0; i < grid.height; i++) {
        for(int j = 0; j < grid.width; j++) {
            if(grid.distance[i][j] < 1000000) {
                int d = grid.tiles[i][j];
                if (i + d < grid.height) {
                    grid.distance[i+d][j] = std::min(grid.distance[i][j] + 1, grid.distance[i+d][j]);
                }
                if (j + d < grid.width) {
                    grid.distance[i][j+d] = std::min(grid.distance[i][j] + 1, grid.distance[i][j+d]);
                }
            }
        }
    }
    if(grid.distance[grid.height-1][grid.width-1] == 1000000) grid.distance[grid.height-1][grid.width-1] = 0;
    std::cout << grid.distance[grid.height-1][grid.width-1] << std::endl;
    //system("pause");
    return 0;
}

Answer 1

需要构造图，这可以通过对矩阵进行一次扫描的动态规划轻松解决。

您可以在开始时将距离矩阵 D[i,j] 设置为 +inf，其中 D[0,0] = 0。在遍历矩阵时，您只需执行

if (D[i,j] < +inf) {
  int d = a[i, j];
  if (i + d < M) {
    D[i + d, j] = min(D[i,j] + 1, D[i + d, j]);
  }
  if (j + d < N) {
    D[i, j + d] = min(D[i,j] + 1, D[i, j + d]);
  }
}

最终的最小距离在D[M -1, N-1]中。如果您希望重建路径，您可以保留一个单独的矩阵来标记最短路径的来源。