leetcode490题，迷宫。
leetcode499题，迷宫III，求到某个点的最小路径“lul”
带返回值与普通的C算法-DFS（深度优先搜索法）在写法上要注意的是：标记访问这个一定要有，如果本身条件不答应，就建立个数组vistied[MAXLEN][MAXLEN]
1、for循环节点，和判断什么时候递归，放在主函数，与回溯风格明显不同。
2、dfs结构：a、if 出界 xxx；b、if被访问过 xxx；c、如果成功 xxx；d、标记访问； c、dfs(next)
3、注意满足条件这个地方一定要好好填
4、新增注意，返回值都要填
比如499题目要传入最小路径做对比，那么需要在如果成功xxxx中进行算法分析。
像这种要增加字符串的，可以把字符串，当前字符串，当前字符串index作为全局变量操作
char g_ret[1000];
char g_final[1000];
int g_c;
bool dfs(int** maze, int row, int col, int sx, int sy, int dx, int dy, int count)
{
if (maze[sx][sy] != 0 && maze[sx][sy] < count) {
return false;
}
if (sx == dx && sy == dy) {
g_ret[g_c++] = '\0';
if (maze[sx][sy] == count && strcmp(g_final, g_ret) < 0) {
g_c--;
return false;
}
maze[sx][sy] = count;
memcpy(g_final, g_ret, g_c);
g_c--;
return true;
}
maze[sx][sy] = count;
int d, l, r, u, dcount, lcount, rcount, ucount;
bool dflag, lflag, rflag, uflag;
d = u = sx;
l = r = sy;
lcount = dcount = rcount = ucount = count;
dflag = lflag = rflag = uflag = false;
while (d + 1 < row && maze[d + 1][sy] != 1) {
d++;
dcount++;
if (sy == dy && d == dx) {
break;
}
}
if (dcount > count) {
g_ret[g_c++] = 'd';
//printf("old sx %d sy %d new sx %d sy %d count %d \n", sx, sy, d, sy, count);
dflag = dfs(maze, row, col, d, sy, dx, dy, dcount);
g_c--;
}
while (l - 1 >= 0 && maze[sx][l - 1] != 1) {
l--;
lcount++;
if (sx == dx && l == dy) {
break;
}
}
if (lcount > count) {
g_ret[g_c++] = 'l';
//printf("old sx %d sy %d new sx %d sy %d count %d \n", sx, sy, sx, l, count);
lflag = dfs(maze, row, col, sx, l, dx, dy, lcount);
g_c--;
}
while (r + 1 < col && maze[sx][r + 1] != 1) {
r++;
rcount++;
if (sx == dx && r == dy) {
break;
}
}
if (rcount > count) {
g_ret[g_c++] = 'r';
//printf("old sx %d sy %d new sx %d sy %d count %d \n", sx, sy, sx, r, count);
rflag = dfs(maze, row, col, sx, r, dx, dy, rcount);
g_c--;
}
while (u - 1 >= 0 && maze[u - 1][sy] != 1) {
u--;
ucount++;
if (sy == dy && u == dx) {
break;
}
}
if (ucount > count) {
g_ret[g_c++] = 'u';
//printf("old sx %d sy %d new sx %d sy %d count %d \n", sx, sy, u, sy, count);
uflag = dfs(maze, row, col, u, sy, dx, dy, ucount);
g_c--;
}
return dflag || rflag || lflag || uflag;
}
char * findShortestWay(int** maze, int mazeSize, int* mazeColSize, int* ball, int ballSize, int* hole, int holeSize) {
int row, col, dx, dy, count;
if (mazeSize == 0) {
return "impossible";
}
if (ball[0] == hole[0] && ball[1] == hole[1]) {
return "impossible";
}
row = mazeSize;
col = *mazeColSize;
dx = hole[0];
dy = hole[1];
count = 2;
g_final[0] = '\0';
g_ret[0] = '\0';
g_c = 0;
if (dfs(maze, row, col, ball[0], ball[1], dx, dy, count) == false) {
return "impossible";
}
int retlen = strlen(g_final) + 1;
char * ret = (char*)malloc(retlen);
memcpy(ret, g_final, retlen);
return ret;
}
490迷宫答案
#define MAXLEN 2000
int visited[MAXLEN][MAXLEN];
bool dfs(int sx, int sy, int dx, int dy, int row, int col, int** maze)
{
if (sx == dx && sy == dy) {
return true;
}
if (sx < 0 || sx >= row || sy < 0 || sy >= col) {
return false;
}
if (visited[sx][sy] == 1) {
return false;
}
//printf("sx %d sy %d\n", sx, sy);
visited[sx][sy] = 1;
int tmpx, tmpy;
tmpx = sx;
tmpy = sy;
while (sx - 1 >= 0 && maze[sx - 1][sy] == 0) {
sx--;
}
if (dfs(sx, sy, dx, dy, row, col, maze) == 1) {
return true;
}
sx = tmpx;
while (sx + 1 < row && maze[sx + 1][sy] == 0) {
sx++;
}
if (dfs(sx, sy, dx, dy, row, col, maze) == 1) {
return true;
}
sx = tmpx;
while (sy + 1 < col && maze[sx][sy + 1] == 0) {
sy++;
}
if (dfs(sx, sy, dx, dy, row, col, maze) == 1) {
return true;
}
sy = tmpy;
while (sy - 1 >= 0 && maze[sx][sy - 1] == 0) {
sy--;
}
if (dfs(sx, sy, dx, dy, row, col, maze) == 1) {
return true;
}
return false;
}
bool hasPath(int** maze, int mazeSize, int* mazeColSize, int* start, int startSize, int* destination, int destinationSize) {
int row, col, dx, dy, sx, sy;
if (mazeSize == 0) {
return false;
}
if (start[0] == destination[0] && start[1] == destination[1]) {
return true;
}
memset(visited, 0, sizeof(int)*MAXLEN*MAXLEN);
row = mazeSize;
col = *mazeColSize;
dx = destination[0];
dy = destination[1];
sx = start[0];
sy = start[1];
return dfs(sx, sy, dx, dy, row, col, maze);
}