CDOJ 149 解救小Q 搜索 BFS

非常显然的BFS搜索

应注意的问题:
1、到了传送点必须传送,不能略过
2、传送点传送不消耗步数
3、传送点可能会经过两次 (在这里跪了好久orz)
例:
2 10
.#Q#......
L.a......a


#include 
#include 
#include 
#include 
using namespace std;

int n,m,T,ans;
char s[60];
int map[60][60]; bool vis[60][60];
int sx,sy,ex,ey;
int a[5]={1,-1,0,0},b[5]={0,0,1,-1};

struct node_pair
{
	int x1,x2,y1,y2;
}chuan[30];

struct node
{
	int x,y,v;
};

queue  q;

void init()
{
	memset(map,0,sizeof(map));
	memset(vis,0,sizeof(vis));
	ans=0;
	while (!q.empty()) q.pop();
	for (int i=1;i<=26;i++)
	{
		chuan[i].x1=0; chuan[i].x2=0;
		chuan[i].y1=0; chuan[i].y2=0;
	}
}

void bfs(int x,int y)
{
	node t;
	t.x=x; t.y=y; t.v=0; vis[x][y]=true;
	q.push(t);
	while (!q.empty())
	{
		node now=q.front(); q.pop();
		if (now.x==ex && now.y==ey)
		{
			ans=now.v;
			break;
		}
		if (map[now.x][now.y]>0)
		{
			int num=map[now.x][now.y];
			if (now.x==chuan[num].x1 && now.y==chuan[num].y1)
			{
				now.x=chuan[num].x2; now.y=chuan[num].y2;
			}
			else
			{
				now.x=chuan[num].x1; now.y=chuan[num].y1;
			}
			//vis[now.x][now.y]=true;
		}
		//printf("  %d %d %d\n",now.x,now.y,now.v);
		for (int i=0;i<4;i++)
		{
			int nx=now.x+a[i],ny=now.y+b[i];
			if (nx>=1 && nx<=n && ny>=1 && ny<=m && map[nx][ny]>=0 && !vis[nx][ny])
			{
				node tt;
				tt.x=nx; tt.y=ny; tt.v=now.v+1;
				if (map[nx][ny]==0) vis[nx][ny]=true;
				q.push(tt);
			}
		}
	}
}

void db()
{
	for (int i=1;i<=26;i++)
	{
		printf("%d %d %d %d %d\n",i,chuan[i].x1,chuan[i].y1,chuan[i].x2,chuan[i].y2);
	}
}

int main()
{
	scanf("%d",&T);
	for (int tt=1;tt<=T;tt++)
	{
		init();
		scanf("%d%d",&n,&m);
		for (int i=1;i<=n;i++)
		{
			scanf("%s",s);
			for (int j=0;j='a' && s[j]<='z')
				{
					int t=s[j]-'a'+1;
					map[i][j+1]=t;
					if (chuan[t].x1==0)
					{
						chuan[t].x1=i; chuan[t].y1=j+1;
					}
					else
					{
						chuan[t].x2=i; chuan[t].y2=j+1;
					}
				}
				if (s[j]=='L')
				{
					sx=i; sy=j+1;
				}
				if (s[j]=='Q')
				{
					ex=i; ey=j+1;
				}
			}
		}
		bfs(sx,sy);
		if (!ans) printf("-1\n");
		else printf("%d\n",ans);
		//db();
	}
	return 0;
}


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