hdu 1175(BFS) 连连看

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<queue>
 5 #include<climits>
 6 using namespace std;
 7 int map[1005][1005],visit[1005][1005];
 8 struct point {
 9     int x,y;
10     int step,dir;//记录转折次数与方向
11 } ;
12 int dx[4]={0,0,1,-1};
13 int dy[4]={1,-1,0,0};
14 int n,m,ex,ey;
15 int bfs(int sx,int sy)
16 {
17     int i;
18     queue<point>Q;
19     point next,now;
20     now.x=sx;now.y=sy;
21     now.step=0;
22     now.dir=-1;
23     Q.push(now);
24     while (!Q.empty())
25     {
26         now=Q.front();
27         Q.pop();
28         if (now.x==ex&&now.y==ey)
29             return 1;
30         for (i=0;i<4;i++)
31         {
32            next.x=now.x+dx[i];
33            next.y=now.y+dy[i];
34            next.step=now.step;
35            next.dir=i;
36            if (next.dir!=now.dir&&now.dir!=-1)//方向发生变化就加一,但是是第一步的时候并不算是转折
37                 next.step++;
38            if ((next.x<=n&&next.x>=1)&&(next.y<=m&&next.y>=1)&&(map[next.x][next.y]==0||(next.x==ex&&next.y==ey)))
39            {
40                if (next.step<3)
41                {
42                    if (next.step<visit[next.x][next.y]) //让visit数组保存的是到每一点的最小转折次数
43                    {
44                        visit[next.x][next.y]=next.step;
45                        Q.push(next);
46                    }
47                }
48            }
49         }
50     }
51     return 0;
52 }
53 int main()
54 {
55     int i,j,t,sx,sy;
56     while (~scanf("%d %d",&n,&m))
57     {
58         if (n==0&&m==0)
59             break;
60         for (i=1;i<=n;i++){
61             for (j=1;j<=m;j++){
62                 scanf("%d",&map[i][j]);
63             }
64         }
65         scanf("%d",&t);
66         while (t--)
67         {
68             cin>>sx>>sy>>ex>>ey;
69             if (sx==ex&&sy==ey)
70             {
71                 printf("NO\n");
72                 continue;
73             }
74             if (map[sx][sy]==0||map[ex][ey]==0||map[sx][sy]!=map[ex][ey])//这些特殊情况不要漏掉
75             {
76                 printf("NO\n");
77                 continue;
78             }
79             for(i = 1; i <= n; i++)
80             for(j = 1; j <= m; j++)
81             visit[i][j] = 10000; //初始化为一个很大的数
82             if (bfs(sx,sy)==1)
83                 printf("YES\n");
84             else
85                 printf("NO\n");
86         }
87 
88     }
89     return 0;
90 }