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
1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 #include<cstring> 5 #include<cmath> 6 #include<queue> 7 using namespace std; 8 int n,m,t; 9 int d1[4][2]={1,0,0,1,-1,0,0,-1}; 10 int d2[4][2]={-1,0,0,-1,1,0,0,1}; 11 char s[25][25]; 12 int vis[25][25][25][25]; 13 struct node 14 { 15 int x1,y1,s,x2,y2; 16 node(){} 17 node(int xx,int yy,int xxx,int yyy,int ss) 18 { 19 x1=xx;y1=yy;x2=xxx;y2=yyy;s=ss; 20 } 21 }st,ed; 22 void bfs() 23 { 24 queue<node> q; 25 node now,next; 26 while(!q.empty()) q.pop(); 27 q.push(node(st.x1,st.y1,st.x2,st.y2,0)); 28 while(!q.empty()) 29 { 30 now=q.front(); 31 q.pop(); 32 //printf("%d %d %d %d %d ",now.x1,now.y1,now.x2,now.y2,now.s); 33 if(fabs(now.x1-now.x2)+fabs(now.y1-now.y2)<2) 34 { 35 printf("%d ",now.s); 36 return; 37 } 38 for(int i=0;i<4;i++) //肯定有一个棋子是移动的，以这个棋子作为判断依据 39 { 40 next.x1=now.x1+d1[i][0]; 41 next.y1=now.y1+d1[i][1]; 42 next.x2=now.x2+d2[i][0]; 43 next.y2=now.y2+d2[i][1]; 44 if(next.x1<0||next.x1>=n||next.y1<0||next.y1>=m||s[next.x1][next.y1]=='X') continue; //这棋子必须移动 45 if(next.x2<0||next.x2>=n||next.y2<0||next.y2>=m||s[next.x2][next.y2]=='X') 46 { 47 next.x2=now.x2,next.y2=now.y2; 48 } 49 if(vis[next.x1][next.y1][next.x2][next.y2]) continue; 50 vis[next.x1][next.y1][next.x2][next.y2]=1; 51 next.s=now.s+1; 52 q.push(next); 53 } 54 } 55 printf("Bad Luck! "); 56 } 57 int main() 58 { 59 int i,j,k; 60 freopen("1.in","r",stdin); 61 while(scanf("%d%d",&n,&m)!=EOF) 62 { 63 for(i=0;i<n;i++) 64 { 65 scanf("%s",s[i]); 66 for(j=0;j<m;j++) 67 { 68 if(s[i][j]=='Z') st.x1=i,st.y1=j; 69 if(s[i][j]=='S') st.x2=i,st.y2=j; 70 } 71 } 72 memset(vis,0,sizeof(vis)); 73 bfs(); 74 } 75 return 0; 76 }