Eight(poj 1077 A*算法，八数码，vis在出队列的时候不要还原啊，否则死循环啊)
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1 //poj 1077 八数码 2 #include <iostream> 3 #include<cmath> 4 #include<queue> 5 #include<stdio.h> 6 #include<stack> 7 #include<stdlib.h> 8 #include<string.h> 9 #define M 10000000 10 using namespace std; 11 int fac[] = {1,1,2,6,24,120,720,5040,40320,362880}; 12 int vis[100000000] = {0,},b[4][2] = {-1,0,1,0,0,-1,0,1}; // u d l r 13 int pre[M],move[M]; 14 int ca = 0; 15 struct node 16 { 17 int n[10]; 18 int numb; 19 int f; 20 int g; 21 friend bool operator < (node n1, node n2) 22 { 23 return n1.f > n2.f; 24 } 25 }; 26 27 int cantor(int s[]) 28 { 29 int n = 9; 30 int i,j,temp = 0,num = 0; 31 for(i = 0;i<n-1;i++) 32 { 33 temp = 0; 34 for(j = i+1;j<n;j++) 35 { 36 if(s[j]<s[i]) temp++; 37 } 38 num += fac[n-i-1]*temp; 39 } 40 return (num+1); 41 } 42 43 int geth(int ss[]) 44 { 45 int t = 0,xx,yy,xxx,yyy; 46 for(int i = 0;i<9;i++) 47 if(ss[i]!=i+1) 48 { 49 if(ss[i] == 9) continue; 50 xx = i/3; 51 yy = i%3; 52 xxx = ss[i]/3; 53 yyy = ss[i]%3; 54 t += abs(xx - xxx) + abs(yy - yyy); 55 } 56 57 return t; 58 } 59 60 void printt(int sss[]) 61 { 62 stack<char> q; 63 64 int p = 1,kk = cantor(sss),cc = 0; 65 //cout<<kk<<" "<<p<<endl; 66 while(p != kk) 67 { 68 //cout<<ca++<<endl; 69 switch(move[p]) 70 { 71 case 0: q.push('u');break; 72 case 1: q.push('d');break; 73 case 2: q.push('l');break; 74 case 3: q.push('r');break; 75 } 76 p = pre[p]; 77 cc++; 78 // cout<<cc<<" "<<p<<" "<<pre[p]<<endl; //为什么总是 2 105 2 105 重复呢？ 79 } 80 while(!q.empty()) 81 { 82 printf("%c",q.top()); 83 q.pop(); 84 } 85 cout<<endl; 86 } 87 88 89 void bfs(int ss[]) 90 { 91 //cout<<cantor(ss)<<endl; 92 memset(pre,-1,sizeof(pre)); 93 bool flag = false; 94 priority_queue<node> q; 95 while(!q.empty()) q.pop(); 96 97 node car; 98 for(int i = 0;i<9;i++) car.n[i] = ss[i]; 99 car.numb = cantor(car.n); 100 car.f = geth(car.n); 101 car.g = 0; 102 q.push(car); 103 vis[car.numb] = 1; 104 while(!q.empty()&&flag == false) 105 { 106 node now = q.top(); 107 q.pop(); 108 //这里不用vis还原啊！！！ 109 int nowx,nowy,gox,goy,nowi = 0; 110 for(nowi = 0;nowi<9;nowi++) 111 { 112 if(now.n[nowi] == 9) break; 113 } 114 nowx = nowi/3; 115 nowy = nowi%3; 116 for(int i = 0;i<4;i++) 117 { 118 if(nowx+b[i][0]>=0&&nowx+b[i][0]<3&&nowy+b[i][1]>=0&&nowy+b[i][1]<3) 119 { 120 gox = nowx+b[i][0]; 121 goy = nowy+b[i][1]; 122 int goi = gox*3 + goy,ch = 0; 123 node togo; 124 for(int j = 0;j<9;j++) togo.n[j] = now.n[j]; 125 togo.f = now.f; 126 togo.numb = now.numb; 127 togo.g = now.g; 128 129 ch = togo.n[nowi]; 130 togo.n[nowi] = togo.n[goi]; 131 togo.n[goi] = ch; 132 133 togo.numb = cantor(togo.n); 134 135 if(vis[togo.numb] == 0&&togo.numb!=pre[now.numb]) 136 { 137 togo.g = now.g + 1; 138 togo.f = geth(togo.n) + togo.g; 139 q.push(togo);//这里是复制还是换指针？？？ 140 //cout<<togo.n[6]<<togo.n[7]<<togo.n[8]<<" "; 141 ca++; 142 vis[togo.numb] = 1; 143 move[togo.numb] = i; 144 pre[togo.numb] = now.numb; 145 //cout<<togo.numb<<" "<<pre[togo.numb]<<endl; 146 // cout<<move[togo.numb]<<endl; 147 if(togo.numb == 1) 148 { 149 flag = true; 150 // cout<<ca<<" truue "<<togo.numb<<endl; 151 // system("pause"); 152 break; 153 } 154 } 155 } 156 } 157 } 158 if(flag) printt(ss); 159 else cout<<"no"<<endl; 160 } 161 162 163 int main() 164 { 165 int i = 0,j = 0,a[10] = {0}; 166 char s[1000] = {0}; 167 gets(s); 168 for(i = 0;s[i]!='\0';i++) 169 { 170 if(s[i] == 'x') 171 { 172 a[j] = 9; 173 j++; 174 } 175 else if(s[i]>='0'&&s[i]<='9') 176 { 177 a[j] = s[i] - '0'; 178 j++; 179 } 180 } 181 182 // for(i = 0;i<3;i++) 183 //for(j = 0;j<3;j++) 184 // cout<<fk[i][j]<<" "<<endl; 185 bfs(a); 186 return 0; 187 }
Eight(poj 1077 A*算法，vis在出队列的时候不要还原啊，否则死循环啊)
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Time Limit: 1000MS |
Memory Limit: 65536K |
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Total Submissions: 18507 |
Accepted: 8242 |
Special Judge |
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Description
The 15-puzzle has been around for over 100 years; even if you don't know it by that name, you've seen it. It is constructed with 15 sliding tiles, each with a number from 1 to 15 on it, and all packed into a 4 by 4 frame with one tile missing. Let's call the missing tile 'x'; the object of the puzzle is to arrange the tiles so that they are ordered as:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 x
where the only legal operation is to exchange 'x' with one of the tiles with which it shares an edge. As an example, the following sequence of moves solves a slightly scrambled puzzle:
1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
5 6 7 8 5 6 7 8 5 6 7 8 5 6 7 8
9 x 10 12 9 10 x 12 9 10 11 12 9 10 11 12
13 14 11 15 13 14 11 15 13 14 x 15 13 14 15 x
r-> d-> r->
The letters in the previous row indicate which neighbor of the 'x' tile is swapped with the 'x' tile at each step; legal values are 'r','l','u' and 'd', for right, left, up, and down, respectively.
Not all puzzles can be solved; in 1870, a man named Sam Loyd was famous for distributing an unsolvable version of the puzzle, and
frustrating many people. In fact, all you have to do to make a regular puzzle into an unsolvable one is to swap two tiles (not counting the missing 'x' tile, of course).
In this problem, you will write a program for solving the less well-known 8-puzzle, composed of tiles on a three by three
arrangement.
Input
You will receive a description of a configuration of the 8 puzzle. The description is just a list of the tiles in their initial positions, with the rows listed from top to bottom, and the tiles listed from left to right within a row, where the tiles are represented by numbers 1 to 8, plus 'x'. For example, this puzzle
1 2 3
x 4 6
7 5 8
is described by this list:
1 2 3 x 4 6 7 5 8
Output
You will print to standard output either the word ``unsolvable'', if the puzzle has no solution, or a string consisting entirely of the letters 'r', 'l', 'u' and 'd' that describes a series of moves that produce a solution. The string should include no spaces and start at the beginning of the line.
Sample Input
2 3 4 1 5 x 7 6 8
Sample Output
ullddrurdllurdruldr
Source
