我有这个问题，我必须通过向右或向下移动找到NxM网格中从A点（总是左上角）到B点（总是在右下角）的最短路径。听起来很简单，嗯？好吧，这里有一个问题：我只能移动我正坐在的瓷砖上显示的数字。让我来说明一下：
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个步骤，从左上角2个节点向下步行到3个，从那里3个节点从右到1，然后从1个节点向下到目标。
[2] 5 1 2
9 2 5 3
[3] 3 1 [1]
4 8 2 [7]
如果不是最短的路径，我也可以采取这条路线：
[2] 5 [1][2]
9 2 5 3
3 3 1 [1]
4 8 2 [7]
不幸的是，这将是一个高达4个步骤，因此，不符合我的利益。 这应该清楚一点。 现在关于输入。
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 //
我已经考虑过这一点，但是无法找到一个更好的解决方案，而不是将输入的网格简化为未加权（或负权重）的图形，并在其上运行类似dijkstra或A *（或沿着这些线的东西）。嗯......这是我迷路的部分。我实现了一些事情（或者立刻投入到捶打中）。它与dijkstra或A *或任何东西无关;只是直接的广度优先搜索。
#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;
}
我应该也指出，运行时间不能超过1秒，最大网格高度和宽度为1000.所有的图块也是1到1000之间的数字。
感谢。
#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;
}
程序现在打印正确的答案。现在我必须优化（运行时间太大）。有关这一个的任何提示吗？优化是我吮吸的一件事。
最后解决方案似乎只包含很少的代码。我喜欢它越少越好。感谢DejanJovanović的美丽解决方案
#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;
}
答案 0 :(得分：3)
你是在思考它。 :)运行广度优先搜索。解空间是二叉树，其中每个节点分支为“右”或“向下”。从当前点开始，生成下点和右点，将其坐标填入队列，重复直到完成。
不检查，像这样：
queue = [{ x: 0, y: 0, path: [] }] # seed queue with starting point
p = nil
do
raise NoSolutionException if p.empty? # solution space exhausted
p = queue.pop # get next state from the back of the queue
break if p.x == MAX_X - 1 && p.y == MAX_Y - 1 # we found final state
l = grid[p.x][p.y] # leap length
# add right state to the front of the queue
queue.unshift({x: p.x + l, y: p.y, path: p.path + [p] }) if p.x + l <= MAX_X
# add down state to the front of the queue
queue.unshift({x: p.x, y: p.y + l, path: p.path + [p] }) if p.y + l <= MAX_Y
end
puts p.path
丑化为C ++留给读者的练习：p
答案 1 :(得分：3)
需要构建图形，使用矩阵上的一次扫描动态编程可以很容易地解决这个问题。
您可以在开始时将距离矩阵D [i，j]设置为+ inf，D [0,0] = 0.在遍历矩阵时，您只需
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]);
}
}
最终的最小距离是D [M -1，N-1]。如果你想重建路径，你可以保留一个单独的矩阵，标记最短路径的来源。
答案 2 :(得分：2)
构建一个未加权的有向图：
N x M个顶点。在下文中，顶点v对应于网格方v。u跳到方格v，则会有从顶点u到v的圆弧。现在应用从右上角到左下角的最短路径算法。
最后，请注意您实际上并不需要构建图形。您可以根据原始网格简单地实现最短路径算法。
答案 3 :(得分：0)
从蛮力方法开始，让它发挥作用，然后从那里进行优化。蛮力是直截了当的：递归地运行它。采取你的两个动作，递归那些，依此类推。收集所有有效答案并保留最低要求。如果运行时间太长，那么您可以通过各种方式进行优化。例如，某些移动可能无效（因为它们超出了网格的维度）并且可以被消除，等等。保持优化，直到最坏情况输入以所需速度运行。
话虽如此，如果您使用相同的系统和输入，性能要求才有意义，即使这样，也有一些警告。 Big O表示法是一种更好的分析性能的方法，而且它可以指向算法并消除分析的需要。