hdu 2216 bfs

题目大意：两个东西朝相同方向移动
Sample Input
4 4
XXXX
.Z..
.XS.
XXXX
4 4
XXXX
.Z..
.X.S
XXXX
4 4
XXXX
.ZX.
.XS.
XXXX
Sample Output
1
1
Bad Luck!

由于两个棋子必然有一个移动。所以假设其中一个一直移动即可，比较水了

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #include<cstring>
 #include<cmath>
 #include<queue>
 using namespace std;
 int n,m,t;
 int d1[][]={,,,,-,,,-};
 int d2[][]={-,,,-,,,,};
 char s[][];
 int vis[][][][];
 struct node
 {
     int x1,y1,s,x2,y2;
     node(){}
     node(int xx,int yy,int xxx,int yyy,int ss)
     {
         x1=xx;y1=yy;x2=xxx;y2=yyy;s=ss;
     }
 }st,ed;
 void bfs()
 {
     queue<node> q;
     node now,next;
     while(!q.empty())   q.pop();
     q.push(node(st.x1,st.y1,st.x2,st.y2,));
     while(!q.empty())
     {
         now=q.front();
         q.pop();
         //printf("%d %d %d %d %d\n",now.x1,now.y1,now.x2,now.y2,now.s);
         if(fabs(now.x1-now.x2)+fabs(now.y1-now.y2)<)
         {
             printf("%d\n",now.s);
             return;
         }
         for(int i=;i<;i++)            //肯定有一个棋子是移动的，以这个棋子作为判断依据
         {
             next.x1=now.x1+d1[i][];
             next.y1=now.y1+d1[i][];
             next.x2=now.x2+d2[i][];
             next.y2=now.y2+d2[i][];
             if(next.x1<||next.x1>=n||next.y1<||next.y1>=m||s[next.x1][next.y1]=='X') continue;       //这棋子必须移动
             if(next.x2<||next.x2>=n||next.y2<||next.y2>=m||s[next.x2][next.y2]=='X')
             {
                 next.x2=now.x2,next.y2=now.y2;
             }
             if(vis[next.x1][next.y1][next.x2][next.y2]) continue;
             vis[next.x1][next.y1][next.x2][next.y2]=;
             next.s=now.s+;
             q.push(next);
         }
     }
     printf("Bad Luck!\n");
 }
 int main()
 {
     int i,j,k;
     freopen("1.in","r",stdin);
     while(scanf("%d%d",&n,&m)!=EOF)
     {
         for(i=;i<n;i++)
         {
             scanf("%s",s[i]);
             for(j=;j<m;j++)
             {
                 if(s[i][j]=='Z')    st.x1=i,st.y1=j;
                 if(s[i][j]=='S')    st.x2=i,st.y2=j;
             }
         }
         memset(vis,,sizeof(vis));
         bfs();
     }
     return ;
 }