POJ 3083 Children of the Candy Corn dfs+bfs

题目连接http://poj.org/problem?id=3083

题目大意：一看就是基本的迷宫问题，一个是优先往左走，一个优先往右走，还有一个是求最短。

额。。。。好久没写bfs= =...上来bfs没加vis= =。。。。然后dfs又写错，因为是左右思路相同，我是直接复制的，找了好久一直是re= =、你妹妹的。。。

用一个全局变量保存朝向，用一个%4来保存结果

View Code

#include <stdio.h>
#include <string.h>
char map[100][100];
int m,n,sx,sy,dx,dy;
int dir;
int lx[4] = {1,0,-1,0};
int ly[4] = {0,-1,0,1};
int rx[4] = {-1,0,1,0};
int ry[4] = {0,-1,0,1};
int leap = 0;
struct node
{
int x,y,step;
}q[100000];
int is_map(int x,int y)
{
if(x >= 0 &&x < n && y >= 0 && y < m)
return 1;
return 0;
}

int dfs(int is_left,int x,int y,int step)
{
if(is_map(x,y) == 0)
return 0;
if(map[x][y] == '#')
return 0;
if(map[x][y] == 'E')
return step+1;

if(is_left)
{
dir = (dir+3)%4;
while(1)
{
int tx,ty;
tx = x+lx[dir];
ty = y+ly[dir];
leap = dfs(1,tx,ty,step+1);
if(leap)
break;
dir = (dir+1)%4;
}
}
else
{
dir = (dir+3)%4;
while(1)
{
int tx,ty;
tx = x+rx[dir];
ty = y+ry[dir];
leap = dfs(0,tx,ty,step+1);
if(leap > 0)
break;
dir = (dir+1)%4;
}
}
return leap;
}
void bfs()
{
int vis[45][45] = {0};
int f,r,i;
f = r = 0;
q[f].x = sx;
q[r].y = sy;
q[r].step = 1;
r++;
while(f<r)
{
struct node temp;
temp = q[f++];
for(i = 0;i < 4;i++)
{
struct node now;
now.x = temp.x+lx[i];
now.y = temp.y+ly[i];
now.step = temp.step+1;
if(is_map(now.x,now.y) && !vis[now.x][now.y])
{
if(map[now.x][now.y] == 'E')
{
printf("%d\n",now.step);
return;
}
vis[now.x][now.y] = 1;
if(map[now.x][now.y] != '#')
q[r++] = now;
}
}
}

}
int main()
{
int i,j,t,lsum,rsum;
leap = 0;
scanf("%d",&t);
while(t--){

scanf("%d %d",&m,&n);
for(i = 0;i < n;i++)
{
scanf("%s",map[i]);

}
for(i = 0;i < n;i++)
for(j = 0;j < m;j++)
{
if(map[i][j] == 'S')
sx = i,sy = j;
if(map[i][j] == 'E')
dx = i,dy = j;
}

dir = 0;
leap = 0;
lsum = dfs(1,sx,sy,0);
dir = 0;
rsum = dfs(0,sx,sy,0);
printf("%d %d ",lsum,rsum);
bfs();
}
return 0;
}