POJ 3083 Children of the Candy Corn

DFS和BFS的综合应用。分别输出一个迷宫由起点开始，左手优先路径长度，右手优先路径长度，以及最短路径长度。其中，前两个采用dfs，而最后一个采用bfs。注意下标w, h不要搞混，好容易wa...另外，用(i + 3) % 4 和 (i + 1) % 4表示方向，也是参考了别人的代码才理解的，画个图，试几个例子就很好理解了。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cstring>
using namespace std;
int dx[4] = {-1, 0, 1, 0};
int dy[4] = {0, -1, 0, 1};
char maze[45][45];
bool flag[45][45];
int w, h;
int cnt;
bool p;
struct node
{
int x, y, step;
};
void dfsl(int x, int y, int i)
{
if (p || maze[x][y] == 'E')
{
p = true;
return;
}
int cx = x + dx[i];
int cy = y + dy[i];
while (cx < 0 || cx >= h || cy < 0 || cy >= w || maze[cx][cy] == '#')
{
i = (i + 3) % 4;
cx = x + dx[i];
cy = y + dy[i];
}
cnt++;
dfsl(cx, cy, (i + 1) % 4);
}
void dfsr(int x, int y, int i)
{
if (p || maze[x][y] == 'E')
{
p = true;
return;
}
int cx = x + dx[i];
int cy = y + dy[i];
while (cx < 0 || cx >= h || cy < 0 || cy >= w || maze[cx][cy] == '#')
{
i = (i + 1) % 4;
cx = x + dx[i];
cy = y + dy[i];
}
cnt++;
dfsr(cx, cy, (i + 3) % 4);
}
int bfs(int start_x, int start_y)
{
queue<node> q;
node start;
start.x = start_x;
start.y = start_y;
start.step = 1;
q.push(start);
flag[start_x][start_y] = 1;
while (!q.empty())
{
node n = q.front();
q.pop();
for (int i = 0; i < 4; i++)
{
int cx = n.x + dx[i];
int cy = n.y + dy[i];
if (flag[cx][cy] || cx < 0 || cx >= h || cy < 0 || cy >= w || maze[cx][cy] == '#')
continue;
start.x = cx;
start.y = cy;
start.step = n.step + 1;
flag[cx][cy] = 1;
if (maze[cx][cy] == 'E')
return start.step;
q.push(start);
}
}
}
int main()
{
int t;
int start_x, start_y;
scanf("%d", &t);
while (t--)
{
memset(flag, 0, sizeof(flag));
scanf("%d%d", &w, &h);
for (int i = 0; i < h; i++)
{
scanf("%s", &maze[i]);
for (int j = 0; j < w; j++)
{
if (maze[i][j] == 'S')
{
start_x = i;
start_y = j;
}
}
}
cnt = 1, p = false;
dfsl(start_x, start_y, 0);
printf("%d ", cnt);
cnt = 1, p = false;
dfsr(start_x, start_y, 0);
printf("%d ", cnt);
printf("%d\n", bfs(start_x, start_y));
}
}