IDA*算法
Orz HZWER
A*+迭代加深搜索=IDA*
这题的估价相当于一个可行性剪枝，即如果当前走的步数s+未归位的点数>搜索深度k，则剪枝
1 /**************************************************************
2 Problem: 1085
3 User: Tunix
4 Language: C++
5 Result: Accepted
6 Time:1388 ms
7 Memory:1272 kb
8 ****************************************************************/
9
10 //BZOJ 1085
11 #include<cstdio>
12 #include<cstdlib>
13 #include<cstring>
14 #include<iostream>
15 #include<algorithm>
16 #define rep(i,n) for(int i=0;i<n;++i)
17 #define F(i,j,n) for(int i=j;i<=n;++i)
18 #define D(i,j,n) for(int i=j;i>=n;--i)
19 using namespace std;
20
21 int getint(){
22 int v=0,sign=1; char ch=getchar();
23 while(ch<'0'||ch>'9') {if (ch=='-') sign=-1; ch=getchar();}
24 while(ch>='0'&&ch<='9') {v=v*10+ch-'0'; ch=getchar();}
25 return v*=sign;
26 }
27 /*******************tamplate********************/
28 int k;
29 const int ans[5][5]={{1,1,1,1,1},
30 {0,1,1,1,1},
31 {0,0,2,1,1},
32 {0,0,0,0,1},
33 {0,0,0,0,0}};
34 const int fx[8]={1,1,-1,-1,2,2,-2,-2},
35 fy[8]={2,-2,2,-2,1,-1,1,-1};
36
37 bool flag=0;
38 bool judge(int a[5][5]){
39 rep(i,5)
40 rep(j,5)
41 if (ans[i][j]!=a[i][j]) return 0;
42 return 1;
43 }
44 bool eva(int a[5][5],int s){
45 int v=0;
46 rep(i,5)
47 rep(j,5)
48 if (a[i][j]!=ans[i][j]){
49 v++;
50 if(v+s>k) return 0;
51 }
52 return 1;
53 }
54 //v¼´Îª¹À¼Ûº¯Êý£¨Èç¹û¸úans²î¾àÌ«´ó£¬²»¿ÉÄÜÔÚk²½Ö®ÄÚ³ö½â£¬ÔòÉáÆú
55 void search(int s,int a[5][5],int x,int y){
56 if (s==k) {if (judge(a)) flag=1; return;}
57 if (flag==1) return;
58 rep(i,8){
59 int tx=x+fx[i],ty=y+fy[i];
60 if (tx<0||tx>4||ty<0||ty>4) continue;
61 swap(a[x][y],a[tx][ty]);
62 if (eva(a,s)) search(s+1,a,tx,ty);
63 swap(a[x][y],a[tx][ty]);
64 }
65 }
66 int main(){
67 int T=getint();
68 char ch[10];
69 while(T--){
70 int a[5][5],x,y;
71 memset(a,0,sizeof a);
72 rep(i,5){
73 scanf("%s",ch);
74 rep(j,5)
75 if (ch[j]=='*'){ a[i][j]=2;x=i;y=j; }
76 else a[i][j]=ch[j]-'0';
77 }
78 flag=0;
79 for(k=1;k<=15;++k){
80 search(0,a,x,y); if (flag){ printf("%d\n",k);break;}
81 }
82 if (!flag) printf("-1\n");
83 }
84 return 0;
85 }
