15 Puzzle LightOJ - 1121

https://cn.vjudge.net/problem/LightOJ-1121

 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #include<vector>
 using namespace std;
 #define fi first
 #define se second
 #define mp make_pair
 #define pb push_back
 typedef long long ll;
 typedef unsigned long long ull;
 char /*ans[1011],*/now[];int len;
 int x,y,a[][];//xx1[4][4]={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,0};
 int xx2[][]={{,},{,},{,},{,},{,},{,},{,},{,},{,},
     {,},{,},{,},{,},{,},{,},{,}};
 bool fl;
 int T,n,maxd;
 inline int abs1(int x){return x>?x:-x;}
 //估价函数：除0以外，所有数当前位置与目标位置的曼哈顿距离之和
 int hh()
 {
     int i,j,cnt=;
     for(i=;i<;++i)
         for(j=;j<;++j)
             if(a[i][j])
                 cnt+=abs1(i-xx2[a[i][j]][])+abs1(j-xx2[a[i][j]][]);
     return cnt;
 }
 inline void swap1(int &a,int &b)
 {
     int t=a;a=b;b=t;
 }
 //#define swap(a,b) {t=a;a=b;b=t;}
 #define swap swap1
 void dfs(int d,char pre)
 {
     //printf("1t%d\n",d);
     int h=hh();
     if(!h)    {fl=;/*memcpy(ans,now,sizeof(char)*(len+1));*/return;}
     if(d+h>maxd)    return;
     if(x!= && pre!='U')
     {
         //now[++len]='D';
         swap(a[x][y],a[x+][y]);++x;
         dfs(d+,'D');
         if(fl)    {now[++len]='D';return;}
         //--len;
         --x;swap(a[x][y],a[x+][y]);
     }
     if(y!= && pre!='R')
     {
         //now[++len]='L';
         swap(a[x][y-],a[x][y]);--y;
         dfs(d+,'L');
         if(fl)    {now[++len]='L';return;}
         //--len;
         ++y;swap(a[x][y-],a[x][y]);
     }
     if(y!= && pre!='L')
     {
         //now[++len]='R';
         swap(a[x][y+],a[x][y]);++y;
         dfs(d+,'R');
         if(fl)    {now[++len]='R';return;}
         //--len;
         --y;swap(a[x][y+],a[x][y]);
     }
     if(x!= && pre!='D')//不要让操作刚好抵消上一次操作，实测很有效
     {
         //now[++len]='U';
         swap(a[x][y],a[x-][y]);--x;
         dfs(d+,'U');
         if(fl)    {now[++len]='U';return;}
         //--len;
         ++x;swap(a[x][y],a[x-][y]);
     }
 }
 /*
     判无解,
     https://blog.csdn.net/obsorb_knowledge/article/details/79915484
     A=将16个数排成一行，((0,0),(0,1),(0,2),(0,3),(1,0),(1,1),..的顺序)
     删去0，当前状态这么做之后的逆序对数与目标状态奇偶性是否相同
     (相同为1，不同为0)
     B=当前状态0的行号与目标状态的奇偶性是否相同
     有解要求满足：A==B
 */
 bool judge()
 {
     int tmp[],x0,i,j;
     tmp[]=;
     for(i=;i<;++i)
         for(j=;j<;++j)
             if(a[i][j])
                 tmp[++tmp[]]=a[i][j];
             else
                 x0=i;
     //for(i=1;i<=15;++i)
     //    printf("1t%d\n",tmp[i]);
     int a1=;
     for(i=;i<=;++i)
         for(j=;j<i;++j)
             if(tmp[j]>tmp[i])
                 ++a1;
     //printf("2t%d %d\n",a1&1,3-x0);
     return (a1&)==((-x0)&);
 }
 int main()
 {
     int i,j;
     scanf("%d",&T);
     for(int TT=;TT<=T;++TT)
     {
         fl=;len=;
         for(i=;i<;++i)
             for(j=;j<;++j)
             {
                 scanf("%d",&a[i][j]);
                 if(a[i][j]==)    x=i,y=j;
             }
         printf("Case %d: ",TT);
         if(!judge())
         {
             puts("This puzzle is not solvable.");
             continue;
         }
         for(maxd=;maxd<=;++maxd)
         {
             dfs(,);
             //printf("1t%d\n",maxd);
             if(fl)    break;
         }
         //printf("2t%d\n",maxd);
         if(fl)
         {
             for(i=len;i>=;--i)
                 printf("%c",now[i]);
             puts("");
         }
         else
             puts("This puzzle is not solvable.");
     }
     return ;
 }

另有

https://cn.vjudge.net/problem/SCU-1110

https://cn.vjudge.net/problem/UVA-10181