hdu 1043 Eight(康托展开,打表)

  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 typedef long long ll;
  4 typedef long long LL;
  5 
  6 /**
  7     学习:(1)康托展开的使用
  8           (2)路径记录
  9         ? A*
 10         ? 双向bfs
 11 */
 12 
 13 int t[9],s[9];
 14 bool vis[362880 + 10];
 15 char ans[362880 + 10][42];
 16 int fac[]={1,1,2,6,24,120,720,5040,40320};
 17 void cantor(int *s, LL num, int k){//康托展开，把一个数字num展开成一个数组s，k是数组长度
 18     int t;
 19     bool h[k];//0到k-1，表示是否出现过
 20     memset(h, 0, sizeof(h));
 21     for(int i = 0; i < k; i ++){
 22         t = num / fac[k-i-1];
 23         num = num % fac[k-i-1];
 24         for(int j = 0, pos = 0; ; j ++, pos ++){
 25             if(h[pos]) j --;
 26             if(j == t){
 27                 h[pos] = true;
 28                 s[i] = pos + 1;
 29                 break;
 30             }
 31         }
 32     }
 33 }
 34 void inv_cantor(int *s, LL &num, int k){//康托逆展开，把一个数组s换算成一个数字num
 35     int cnt;
 36     num = 0;
 37     for(int i = 0; i < k; i ++){
 38         cnt = 0;
 39         for(int j = i + 1; j < k; j ++){
 40             if(s[i] > s[j]) cnt ++;//判断几个数小于它
 41         }
 42         num += fac[k-i-1] * cnt;
 43     }
 44 }
 45 
 46 struct Node{
 47     int key,cnt,x,y; // x,y 表示 9 所在的位置
 48     Node(int _key,int _x,int _y,int _cnt):key(_key),cnt(_cnt),x(_x),y(_y){}
 49 };
 50 
 51 void init(){
 52 
 53     ll skey;
 54     memset(vis,false,sizeof(vis));
 55     memset(ans,0,sizeof(ans));
 56 
 57     for(int i = 1 ; i <= 9 ; i ++) s[i-1] = i;
 58     inv_cantor(s,skey,9);
 59     vis[skey] = true;
 60     ans[skey][0] = ‘\0‘;
 61      queue<Node> q;
 62     q.push(Node{skey,2,2,0});
 63 
 64     while(!q.empty()){
 65         Node tmp = q.front(); q.pop();
 66         ll nowkey = tmp.key,newkey;
 67         int cnt = tmp.cnt,  nowx = tmp.x , nowy = tmp.y;
 68         int nowarr[10];
 69         cantor(nowarr,nowkey, 9);
 70 
 71         if(nowx >= 1){
 72             swap(nowarr[nowx*3+nowy],nowarr[(nowx-1)*3+nowy]);
 73             inv_cantor(nowarr, newkey, 9);
 74             if(!vis[newkey]){
 75                 vis[newkey] = true;
 76                 for(int i = 0 ; i < cnt ; i ++) ans[newkey][i]=ans[nowkey][i];
 77                 ans[newkey][cnt] = ‘d‘;
 78                 ans[newkey][cnt+1] = ‘\0‘;
 79                 q.push(Node{newkey,nowx-1,nowy,cnt+1});
 80             }
 81             swap(nowarr[nowx*3+nowy],nowarr[(nowx-1)*3+nowy]);
 82         }
 83 
 84         if(nowy >= 1){
 85             swap(nowarr[nowx*3+nowy],nowarr[nowx*3+nowy-1]);
 86             inv_cantor(nowarr, newkey, 9);
 87             if(!vis[newkey]){
 88                 vis[newkey] = true;
 89                 for(int i = 0 ; i < cnt ; i ++) ans[newkey][i]=ans[nowkey][i];
 90                 ans[newkey][cnt] = ‘r‘;
 91                 ans[newkey][cnt+1] = ‘\0‘;
 92                 q.push(Node{newkey,nowx,nowy-1,cnt+1});
 93             }
 94             swap(nowarr[nowx*3+nowy],nowarr[nowx*3+nowy-1]);
 95         }
 96         if(nowx <= 1){
 97             swap(nowarr[nowx*3+nowy],nowarr[(nowx+1)*3+nowy]);
 98             inv_cantor(nowarr, newkey, 9);
 99             if(!vis[newkey]){
100                 vis[newkey] = true;
101                 for(int i = 0 ; i < cnt ; i ++) ans[newkey][i]=ans[nowkey][i];
102                 ans[newkey][cnt] = ‘u‘;
103                 ans[newkey][cnt+1] = ‘\0‘;
104                 q.push(Node{newkey,nowx+1,nowy,cnt+1});
105             }
106             swap(nowarr[nowx*3+nowy],nowarr[(nowx+1)*3+nowy]);
107         }
108         if(nowy <= 1){
109             swap(nowarr[nowx*3+nowy],nowarr[nowx*3+nowy+1]);
110             inv_cantor(nowarr, newkey, 9);
111             if(!vis[newkey]){
112                 vis[newkey] = true;
113                 for(int i = 0 ; i < cnt ; i ++) ans[newkey][i]=ans[nowkey][i];
114                 ans[newkey][cnt] = ‘l‘;
115                 ans[newkey][cnt+1] = ‘\0‘;
116                 q.push(Node{newkey,nowx,nowy+1,cnt+1});
117             }
118             swap(nowarr[nowx*3+nowy],nowarr[nowx*3+nowy+1]);
119         }
120 
121     }
122 }
123 
124 void solve(){
125 
126     ll tkey;
127     inv_cantor(t,tkey,9);
128 
129     if(!vis[tkey]) puts("unsolvable");
130     else{
131         int len = strlen(ans[tkey]);
132         for(int i = len - 1 ; i >= 0 ; i --) putchar(ans[tkey][i]);
133         puts("");
134     }
135 }
136 
137 int main(){
138     string str;
139     char x;
140     int cnt;
141     init();
142     while(getline(cin,str)){
143         stringstream ss(str);
144         int cnt = 0;
145         while(ss >> x){
146             if(x == ‘x‘) t[cnt++] = 9;
147             else t[cnt++] = x - ‘0‘;
148         }
149         solve();
150     }
151     return 0;
152 }