【BZOJ】【1085】【SCOI2005】骑士精神

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

　　

 /**************************************************************
     Problem: 1085
     User: Tunix
     Language: C++
     Result: Accepted
     Time:1388 ms
     Memory:1272 kb
 ****************************************************************/

 //BZOJ 1085
 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 #define rep(i,n) for(int i=0;i<n;++i)
 #define F(i,j,n) for(int i=j;i<=n;++i)
 #define D(i,j,n) for(int i=j;i>=n;--i)
 using namespace std;

 int getint(){
     int v=,sign=; char ch=getchar();
     while(ch<''||ch>'') {if (ch=='-') sign=-; ch=getchar();}
     while(ch>=''&&ch<='') {v=v*+ch-''; ch=getchar();}
     return v*=sign;
 }
 /*******************tamplate********************/
 int k;
 const int ans[][]={{,,,,},
                      {,,,,},
                      {,,,,},
                      {,,,,},
                      {,,,,}};
 const int fx[]={,,-,-,,,-,-},
           fy[]={,-,,-,,-,,-};

 bool flag=;
 bool judge(int a[][]){
     rep(i,)
         rep(j,)
             if (ans[i][j]!=a[i][j]) return ;
     return ;
 }
 bool eva(int a[][],int s){
     int v=;
     rep(i,)
         rep(j,)
             if (a[i][j]!=ans[i][j]){
                 v++;
                 if(v+s>k) return ;
             }
     return ;
 }
 //v¼´Îª¹À¼Ûº¯Êý£¨Èç¹û¸úans²î¾àÌ«´ó£¬²»¿ÉÄÜÔÚk²½Ö®ÄÚ³ö½â£¬ÔòÉáÆú
 void search(int s,int a[][],int x,int y){
     if (s==k) {if (judge(a)) flag=; return;}
     if (flag==) return;
     rep(i,){
         int tx=x+fx[i],ty=y+fy[i];
         if (tx<||tx>||ty<||ty>) continue;
         swap(a[x][y],a[tx][ty]);
         if (eva(a,s)) search(s+,a,tx,ty);
         swap(a[x][y],a[tx][ty]);
     }
 }
 int main(){
     int T=getint();
     char ch[];
     while(T--){
         int a[][],x,y;
         memset(a,,sizeof a);
         rep(i,){
             scanf("%s",ch);
             rep(j,)
                 if (ch[j]=='*'){ a[i][j]=;x=i;y=j; }
                 else a[i][j]=ch[j]-'';
         }
         flag=;
         for(k=;k<=;++k){
             search(,a,x,y); if (flag){ printf("%d\n",k);break;}
         }
         if (!flag) printf("-1\n");
     }
     return ;
 }