HDU 2612 Find a way（BFS）

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<queue>
 4 #include<algorithm>
 5 using namespace std;
 6 const int N=205;
 7 
 8 int n,m,ans,cnt;
 9 int d[4][2]={{0,1},{1,0},{0,-1},{-1,0}};
10 int vis[N][N],resy[N][N],resm[N][N];
11 char map[N][N];
12 
13 struct node{
14     int x,y,step;
15 }pre,now;
16 
17 struct node2{
18     int x,y;    
19 }kfc[N*N];
20 
21 void bfs(int x,int y,int name){
22     queue<node>q;
23     now.x=x;
24     now.y=y;
25     now.step=0;
26     q.push(now);
27     int num=0;
28     while(!q.empty()){
29         pre=q.front();
30         q.pop();
31         for(int i=0;i<4;i++){
32             int xx=pre.x+d[i][0];
33             int yy=pre.y+d[i][1];
34             if(xx<1||yy<1||xx>n||yy>m||map[xx][yy]==‘#‘||vis[xx][yy])
35                 continue;
36             //打表记录Y、M到各个@点的最小距离 
37             if(map[xx][yy]==‘@‘){
38                 num++;
39                 if(name==1)
40                     resy[xx][yy]=pre.step+1;
41                 else
42                     resm[xx][yy]=pre.step+1;
43                 if(num==cnt)
44                     return;
45             }
46             vis[xx][yy]=1;
47             now.x=xx;
48             now.y=yy;
49             now.step=pre.step+1;
50             q.push(now);
51         }
52     }
53     return;
54 }
55 
56 int main(){
57     while(~scanf("%d%d",&n,&m)){
58         memset(vis,0,sizeof(vis));
59         memset(resy,0x3f,sizeof(resy));
60         memset(resm,0x3f,sizeof(resm));
61         int sx,sy,sxx,syy;
62         cnt=0;
63         for(int i=1;i<=n;i++){
64             getchar();
65             for(int j=1;j<=m;j++){
66                 scanf("%c",&map[i][j]);
67                 if(map[i][j]==‘@‘)
68                     kfc[++cnt].x=i,kfc[cnt].y=j;
69                 if(map[i][j]==‘Y‘)
70                     sx=i,sy=j;
71                 if(map[i][j]==‘M‘)
72                     sxx=i,syy=j;
73             }
74         }
75         bfs(sx,sy,1);
76         memset(vis,0,sizeof(vis));
77         bfs(sxx,syy,2);
78         int ans=1<<30;
79         for(int i=1;i<=cnt;i++){
80             int x=kfc[i].x;
81             int y=kfc[i].y;
82             ans=min(ans,resy[x][y]+resm[x][y]);
83         }
84         printf("%d\n",ans*11);
85     }
86     return 0;
87 }