我遇到这个问题，我必须通过向右或向下移动来找到 NxM 网格中从 A 点（总是左上角）到 B 点（总是右下角）的最短路径。听起来很容易，是吗？问题是：我只能移动我现在坐在的图块上显示的数字。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 个节点，从那里向右 3 个节点走到 1 个节点，然后向下 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，这需要一个多达4步，因此，不符合我的利益。
这应该会让事情变得更清楚一些。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 //
Homework
我已经考虑过这一点，但无法找到比将输入的网格简化为未加权（或负权重）图表并在其上运行类似 dijkstra 或 A*（或类似的东西）的更好的解决方案。嗯...这是我迷路的部分。我一开始就实现了一些东西（或者立即扔掉一些东西）。它与 dijkstra 或 A* 或任何东西都没有关系；只是直接的广度优先搜索。
The Code
#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;
}
我或许还应该指出，运行时间不能超过 1 秒，最大网格高度和宽度为 1000。所有图块也是从 1 到 1000 的数字。
Thanks.
编辑后的代码
#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的答案。现在我必须优化（运行时间太大）。关于这个有什么提示吗？优化是我最讨厌的一件事。
答案
最后，该解决方案似乎只包含很少的代码。越少越好，因为我喜欢。感谢 Dejan Jovanović 提供了完美的解决方案
#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;
}
