迷宫求解算法:深度优先搜索
#include 'maze.h'
int map[row][col] = { {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}, {1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1}, {1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1}, {1, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1}, {1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 1}, {1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1}, {1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1}, {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1} }; // 迷宫地图,1代表墙壁,0代表路
int enterx = 1, entery = 1; // 入口坐标 int exitx = 6, exity = 9; // 出口坐标
Path path[(row - 1) * (col - 1)]; // 探索迷宫的路径 int pathLength = 0; // 路径长度
int main(int argc, char* argv[]) { LinkStack L; L.count = 0; CreateLink(&L);
bool flag = Maze(&L);
if (flag == true) {
PrintPath();
} else {
printf("Can not find path");
}
return 0;
}
// 判断是否是走过的迷宫块 bool Panduan(int x, int y) { if (map[x][y] == 0) { return true; }
return false;
}
// 探索迷宫可通行的路径 bool Maze(LinkStack* L) { bool flag = false;
LinkStackPtr p;
p = (LinkStackPtr)malloc(sizeof(struct StackNode));
p->position.x = enterx;
p->position.y = entery;
p->direct.top = 1;
p->direct.right = 0;
p->direct.left = 0;
p->direct.bottom = 0;
map[p->position.x][p->position.y] = 2;
// 将入口结点入栈
Push(L, p);
while (!IsEmpty(L)) {
LinkStackPtr top = GetTop(L);
if (top->position.x == exitx && top->position.y == exity) {
flag = true;
break;
}
bool found = false;
// 向上探索
if (top->direct.top && top->position.x - 1 >= 0 && Panduan(top->position.x - 1, top->position.y)) {
LinkStackPtr newNode = (LinkStackPtr)malloc(sizeof(struct StackNode));
newNode->position.x = top->position.x - 1;
newNode->position.y = top->position.y;
newNode->direct.top = 1;
newNode->direct.right = 0;
newNode->direct.left = 0;
newNode->direct.bottom = 0;
map[newNode->position.x][newNode->position.y] = 2;
Push(L, newNode);
RecordExplorationPath(newNode);
found = true;
continue;
}
// 向右探索
if (top->direct.right && top->position.y + 1 < col && Panduan(top->position.x, top->position.y + 1)) {
LinkStackPtr newNode = (LinkStackPtr)malloc(sizeof(struct StackNode));
newNode->position.x = top->position.x;
newNode->position.y = top->position.y + 1;
newNode->direct.top = 0;
newNode->direct.right = 1;
newNode->direct.left = 0;
newNode->direct.bottom = 0;
map[newNode->position.x][newNode->position.y] = 2;
Push(L, newNode);
RecordExplorationPath(newNode);
found = true;
continue;
}
// 向左探索
if (top->direct.left && top->position.y - 1 >= 0 && Panduan(top->position.x, top->position.y - 1)) {
LinkStackPtr newNode = (LinkStackPtr)malloc(sizeof(struct StackNode));
newNode->position.x = top->position.x;
newNode->position.y = top->position.y - 1;
newNode->direct.top = 0;
newNode->direct.right = 0;
newNode->direct.left = 1;
newNode->direct.bottom = 0;
map[newNode->position.x][newNode->position.y] = 2;
Push(L, newNode);
RecordExplorationPath(newNode);
found = true;
continue;
}
// 向下探索
if (top->direct.bottom && top->position.x + 1 < row && Panduan(top->position.x + 1, top->position.y)) {
LinkStackPtr newNode = (LinkStackPtr)malloc(sizeof(struct StackNode));
newNode->position.x = top->position.x + 1;
newNode->position.y = top->position.y;
newNode->direct.top = 0;
newNode->direct.right = 0;
newNode->direct.left = 0;
newNode->direct.bottom = 1;
map[newNode->position.x][newNode->position.y] = 2;
Push(L, newNode);
RecordExplorationPath(newNode);
found = true;
continue;
}
if (!found) {
map[top->position.x][top->position.y] = 3;
Pop(L);
}
}
return flag;
}
// 判断链栈是否为空 bool IsEmpty(LinkStack *L) { if (L->count == 0) return true; return false; }
// 创建链栈 void CreateLink(LinkStack *L) { L->top = (LinkStackPtr)malloc(sizeof(struct StackNode)); if (!L->top) { exit(1); } L->top = NULL; L->count = 0; }
// 获取栈顶 LinkStackPtr GetTop(LinkStack *L) { return L->top; }
// 进栈 void Push(LinkStack *L, LinkStackPtr s) { s->next = L->top; L->top = s; L->count++; }
// 出栈 void Pop(LinkStack *L) { LinkStackPtr p; if (IsEmpty(L)) { return; } p = L->top; L->top = L->top->next; free(p); L->count--; }
// 记录探索路径 void RecordExplorationPath(LinkStackPtr m){ path[pathLength].x = m->position.x; path[pathLength].y = m->position.y; path[pathLength].seq = pathLength + 1; pathLength += 1; }
// 打印迷宫可通行的路径 void PrintPath(){ for (int i = 0; i < pathLength; i++) { printf("%d\t%d:(%d, %d)\n", path[i].seq, i+1, path[i].x, path[i].y); } return;
原文地址: https://www.cveoy.top/t/topic/peME 著作权归作者所有。请勿转载和采集!