[PKU 1915 2243] 搜索之BFS & A*(续)
{
其实 能用BFS和A*解决的搜索问题很多
比如有一个很simple的骑士巡游问题
就可以这么做
本文只是贴个代码 狗尾续貂 扩充一下
以后有类似的问题也会在这篇文章里补充
}
原题 http://acm.pku.edu.cn/JudgeOnline/problem?id=1915
//2243与这个题基本类似 注意读入输出的细节
用裸BFS 双向BFS A*皆可
如果用A*
应该自己试着去构造一些启发函数
多多尝试 总有一个最好的
我用的就是很简单的曼哈顿距离作启发
效果不错 值得推荐
但不否认有更好的启发函数 下面贴代码
裸BFS
BFS
1 const maxq=100000;
2 maxs=300;
3 dx:array[1..8]of longint=(1,2,2,1,-1,-2,-2,-1);
4 dy:array[1..8]of longint=(2,1,-1,-2,-2,-1,1,2);
5 var n,s,i,gx,gy,t,h,tx,ty:longint;
6 x,y,dep:array[1..maxq]of longint;
7 hash:array[0..maxs-1,0..maxs-1]of longint;
8 procedure main;
9 var i:longint;
10 begin
11 readln(s);
12 readln(x[1],y[1]);
13 readln(gx,gy);
14 if (x[1]=gx)and(y[1]=gy)
15 then begin
16 writeln(0);
17 exit;
18 end;
19 fillchar(hash,sizeof(hash),0);
20 hash[x[1],y[1]]:=1;
21 t:=1; h:=1; dep[0]:=0;
22 while h<=t do
23 begin
24 for i:=1 to 8 do
25 begin
26 tx:=x[h]+dx[i];
27 ty:=y[h]+dy[i];
28 if (tx=gx)and(ty=gy)
29 then begin
30 writeln(dep[h]+1);
31 exit;
32 end;
33 if (tx>=0)and(tx<=s-1)and(ty>=0)and(ty<=s-1)and(hash[tx,ty]=0)
34 then begin
35 hash[tx,ty]:=1;
36 inc(t);
37 dep[t]:=dep[h]+1;
38 x[t]:=tx; y[t]:=ty;
39 end;
40 end;
41 inc(h);
42 end;
43 end;
44 begin
45 assign(input,'knight.in'); reset(input);
46 assign(output,'knight.out'); rewrite(output);
47 readln(n);
48 for i:=1 to n do main;
49 close(input); close(output);
50 end.
51
双向BFS
BFS-双向
1 const maxq=50000;
2 maxs=300;
3 dx:array[1..8]of longint=(1,2,2,1,-1,-2,-2,-1);
4 dy:array[1..8]of longint=(2,1,-1,-2,-2,-1,1,2);
5 var x,y,dep:array[1..maxq,1..2]of longint;
6 hash,kind:array[0..maxs-1,0..maxs-1]of longint;
7 n,s,i,h1,t1,h2,t2,tx,ty:longint;
8 procedure main;
9 var i:longint;
10 begin
11 readln(s);
12 readln(x[1,1],y[1,1]);
13 readln(x[1,2],y[1,2]);
14 if (x[1,1]=x[1,2])and(y[1,1]=y[1,2])
15 then begin
16 writeln(0);
17 exit;
18 end;
19 fillchar(hash,sizeof(hash),0);
20 hash[x[1,1],y[1,1]]:=1; hash[x[1,2],y[1,2]]:=1;
21 kind[x[1,1],y[1,1]]:=1; kind[x[1,2],y[1,2]]:=2;
22 h1:=1; t1:=1; h2:=1; t2:=1;
23 dep[1,1]:=0; dep[1,2]:=0;
24 while (h1<=t1)and(h2<=t2) do
25 begin
26 if dep[t1,1]<dep[t2,2]
27 then begin
28 for i:=1 to 8 do
29 begin
30 tx:=x[h1,1]+dx[i];
31 ty:=y[h1,1]+dy[i];
32 if (tx<0)or(tx>s-1)or(ty<0)or(ty>s-1) then continue;
33 if hash[tx,ty]<>0
34 then begin
35 if kind[tx,ty]=2
36 then begin
37 writeln(dep[h1,1]+1+dep[hash[tx,ty],2]);
38 exit;
39 end;
40 continue;
41 end;
42 inc(t1);
43 x[t1,1]:=tx; y[t1,1]:=ty;
44 hash[tx,ty]:=t1;
45 kind[tx,ty]:=1;
46 dep[t1,1]:=dep[h1,1]+1;
47 end;
48 inc(h1);
49 end
50 else begin
51 for i:=1 to 8 do
52 begin
53 tx:=x[h2,2]+dx[i];
54 ty:=y[h2,2]+dy[i];
55 if (tx<0)or(tx>s-1)or(ty<0)or(ty>s-1) then continue;
56 if hash[tx,ty]<>0
57 then begin
58 if kind[tx,ty]=1
59 then begin
60 writeln(dep[h2,2]+1+dep[hash[tx,ty],1]);
61 exit;
62 end;
63 continue;
64 end;
65 inc(t2);
66 x[t2,2]:=tx; y[t2,2]:=ty;
67 hash[tx,ty]:=t2;
68 kind[tx,ty]:=2;
69 dep[t2,2]:=dep[h2,2]+1;
70 end;
71 inc(h2);
72 end;
73 end;
74 end;
75 begin
76 assign(input,'knight.in'); reset(input);
77 assign(output,'knight.out'); rewrite(output);
78 readln(n);
79 for i:=1 to n do main;
80 close(input); close(output);
81 end.
82
A*
A*
1 const maxq=100000;
2 maxs=300;
3 dx:array[1..8]of longint=(1,2,2,1,-1,-2,-2,-1);
4 dy:array[1..8]of longint=(2,1,-1,-2,-2,-1,1,2);
5 var x,y,dep,h,f:array[1..maxq]of longint;
6 hash:array[0..maxs-1,0..maxs-1]of longint;
7 n,s,i,gx,gy,hs,qs,tx,ty:longint;
8 procedure main;
9 var i,j,now,temp:longint;
10 begin
11 readln(s);
12 readln(x[1],y[1]);
13 readln(gx,gy);
14 if (x[1]=gx)and(y[1]=gy)
15 then begin
16 writeln(0);
17 exit;
18 end;
19 fillchar(hash,sizeof(hash),0);
20 hash[x[1],y[1]]:=1;
21 hs:=1; h[1]:=1; f[1]:=(abs(gx-x[1])+abs(gy-y[1]))div 3;
22 qs:=1; dep[1]:=0;
23 while hs>0 do
24 begin
25 now:=h[1];
26 if (x[now]=gx)and(y[now]=gy)
27 then begin
28 writeln(dep[now]);
29 exit;
30 end;
31 h[1]:=h[hs]; f[1]:=f[hs];
32 dec(hs);
33 i:=1;
34 while true do
35 begin
36 if i shl 1>hs then break;
37 if (i shl 1+1>hs)or(f[i shl 1]<f[i shl 1+1])
38 then begin
39 if f[i shl 1]<f[i]
40 then begin
41 temp:=h[i shl 1]; h[i shl 1]:=h[i]; h[i]:=temp;
42 temp:=f[i shl 1]; f[i shl 1]:=f[i]; f[i]:=temp;
43 i:=i shl 1;
44 end
45 else break;
46 end
47 else begin
48 if f[i shl 1+1]<f[i]
49 then begin
50 temp:=h[i shl 1+1]; h[i shl 1+1]:=h[i]; h[i]:=temp;
51 temp:=f[i shl 1+1]; f[i shl 1+1]:=f[i]; f[i]:=temp;
52 i:=i shl 1+1;
53 end
54 else break;
55 end;
56 end;
57 for i:=1 to 8 do
58 begin
59 tx:=x[now]+dx[i];
60 ty:=y[now]+dy[i];
61 if (tx<0)or(tx>s-1)or(ty<0)or(ty>s-1)or(hash[tx,ty]>0)and(dep[hash[tx,ty]]<=dep[now]+1) then continue;
62 inc(qs);
63 x[qs]:=tx; y[qs]:=ty;
64 dep[qs]:=dep[now]+1;
65 hash[tx,ty]:=qs;
66 inc(hs);
67 h[hs]:=qs;
68 f[hs]:=dep[qs]+(abs(gx-x[qs])+abs(gy-y[qs]))div 3;
69 j:=hs;
70 while j>1 do
71 begin
72 if f[j shr 1]>f[j]
73 then begin
74 temp:=h[j shr 1]; h[j shr 1]:=h[j]; h[j]:=temp;
75 temp:=f[j shr 1]; f[j shr 1]:=f[j]; f[j]:=temp;
76 j:=j shr 1;
77 end
78 else break;
79 end;
80 end;
81 end;
82 end;
83 begin
84 assign(input,'knight.in'); reset(input);
85 assign(output,'knight0.out'); rewrite(output);
86 readln(n);
87 for i:=1 to n do main;
88 close(input); close(output);
89 end.
90
(未完待续)
BOB HAN 原创 转载请注明出处 http://www.cnblogs.com/Booble/
posted on 2010-09-10 21:34 Master_Chivu 阅读(512) 评论(0) 编辑 收藏 举报
