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 <cstdio>#include <cstring>#include <algorithm>#include <queue>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 <node> 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<m;j++){if (s[j]=='#') map[i][j+1]=-1;if (s[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;}