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 ",k);break;} 81 } 82 if (!flag) printf("-1 "); 83 } 84 return 0; 85 }