jps算法c++代码

当然，我可以为您提供JPS算法的C代码示例。下面是一个简单的实现：

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

#define ROWS 10
#define COLS 10

typedef struct {
    int x;
    int y;
} Point;

typedef struct {
    int f;
    int g;
    int h;
    Point parent;
    Point current;
} Node;

int heuristic(Point start, Point end) {
    return abs(end.x - start.x) + abs(end.y - start.y);
}

bool isValid(int x, int y) {
    return (x >= 0 && x < ROWS && y >= 0 && y < COLS);
}

bool isBlocked(int grid[ROWS][COLS], int x, int y) {
    return (grid[x][y] == 1);
}

bool isDestination(Point current, Point end) {
    return (current.x == end.x && current.y == end.y);
}

void tracePath(Node node, int grid[ROWS][COLS]) {
    printf("Path: \n");
    while (!(node.parent.x == node.current.x && node.parent.y == node.current.y)) {
        printf("(%d, %d)\n", node.current.x, node.current.y);
        Point temp = node.parent;
        node.parent = node.parent.parent;
        node.current = temp;
    }
    printf("(%d, %d)\n", node.current.x, node.current.y);
}

void jumpPointSearch(int grid[ROWS][COLS], Point start, Point end) {
    if (!isValid(start.x, start.y) || !isValid(end.x, end.y)) {
        printf("Invalid start or end point.\n");
        return;
    }
    
    if (isBlocked(grid, start.x, start.y) || isBlocked(grid, end.x, end.y)) {
        printf("Start or end point is blocked.\n");
        return;
    }
    
    bool closedList[ROWS][COLS];
    memset(closedList, false, sizeof(closedList));
    
    Node node;
    node.f = 0;
    node.g = 0;
    node.h = 0;
    node.parent = start;
    node.current = start;
    
    int dx[] = {-1, 1, 0, 0, -1, -1, 1, 1};
    int dy[] = {0, 0, -1, 1, -1, 1, -1, 1};

    while (!isDestination(node.current, end)) {
        int x = node.current.x;
        int y = node.current.y;
        
        closedList[x][y] = true;
        
        for (int i = 0; i < 8; i++) {
            int newX = x + dx[i];
            int newY = y + dy[i];
            
            if (isValid(newX, newY) && !closedList[newX][newY] && !isBlocked(grid, newX, newY)) {
                int gNew = node.g + abs(dx[i]) + abs(dy[i]);
                int hNew = heuristic({newX, newY}, end);
                int fNew = gNew + hNew;
                
                if (grid[newX][newY] == 0 || fNew < grid[newX][newY]) {
                    grid[newX][newY] = fNew;
                    
                    Node newNode;
                    newNode.f = fNew;
                    newNode.g = gNew;
                    newNode.h = hNew;
                    newNode.parent = node.current;
                    newNode.current = {newX, newY};
                    
                    node = newNode;
                }
            }
        }
    }
    
    tracePath(node, grid);
}

int main() {
    int grid[ROWS][COLS] = {
        {0, 0, 1, 0, 0, 0, 0, 0, 0, 0},
        {0, 0, 1, 0, 0, 0, 0, 0, 0, 0},
        {0, 0, 1, 0, 0, 0, 0, 0, 0, 0},
        {0, 0, 1, 0, 0, 0, 0, 0, 0, 0},
        {0, 0, 1, 0, 1, 1, 1, 1, 1, 1},
        {0, 0, 1, 0, 1, 0, 0, 0, 0, 1},
        {0, 0, 1, 0, 1, 1, 1, 1, 0, 1},
        {0, 0, 1, 1, 1, 1, 1, 1, 0, 1},
        {0, 0, 0, 0, 1, 0, 1, 1, 1, 1},
        {0, 0, 1 ,1 ,1 ,1 ,1 ,1 ,1 ,1}
    };
    
    Point start = {0 ,2};
    Point end = {9 ,9};
    
    jumpPointSearch(grid, start ,end);
    
    return 0;
}

这是一个简单的网格地图上使用JPS算法寻找路径的示例。请注意，这个示例是针对一个10x10的网格地图，您可以根据需要进行适当的调整。在代码中，0表示可通过的区域，1表示阻塞的区域。算法将打印出从起点到终点的路径坐标。