Knights in FEN UVA10422 - c++编程基础

Problem D

Knights in FEN

Input: standard input

Output: standard output

Time Limit: 10 seconds

There are black and white knights on a 5 by 5 chessboard. There are twelve of each color, and there is one square that is empty. At any time, a knight can move into an empty square as long as it moves like a knight in normal chess (what else did you expect ).

Given an initial position of the board, the question is: what is the minimum number of moves in which we can reach the final position which is:

Input< http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">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"206" height="207" src="https://www.cppentry.com/upload_files/article/49/1_yhdrh__.jpg" alt="\">

Output

For each set your task is to find the minimum number of moves leading from the starting input configuration to the final one. If that number is bigger than 10, then output one line stating

Unsolvable in less than 11 move(s).

otherwise output one line stating

Solvable in n move(s).

where n <= 10.

The output for each set is produced in a single line as shown in the sample output.

Sample Input

2

01011

110 1

01110

01010

00100

10110

01 11

10111

01001

00000

Sample Output

Unsolvable in less than 11 move(s).

Solvable in 7 move(s).

(Problem Setter: Piotr Rudnicki, University of Alberta, Canada)

“A man is as great as his dreams.”

题意：给出一个状态图，移动其中的骑士恢复到初始的状态。

典型的隐式图搜索问题，BFS搜索+哈希，哈希我是用的set实现的，另外要注意的是国际象棋中骑士走的是日字。

#include #include #include #include using namespace std; typedef int state[25]; const int maxn=5000000; const int dx[]={1,2,2,1,-1,-2,-2,-1}; const int dy[]={-2,-1,1,2,2,1,-1,-2}; state st[maxn]; int dist[maxn]; int front,rear,s; set vis; state goal={1,1,1,1,1,0,1,1,1,1,0,0,2,1,1,0,0,0,0,1,0,0,0,0,0}; int try_to_insert(int s)//哈希函数 { int v=0; for(int i=0;i<25;i++) v=v*2+st[s][i]; if(vis.count(v))return 0; vis.insert(v); return 1; } int bfs() { front=1,rear=2; vis.clear(); while(front 10) return -1;//剪枝，超过10步认为不可达 if(memcmp(goal,p,sizeof(p))==0) return front; int i,j,x,y; for(i=0;i<25;i++) if(st[front][i]==2) break; int z=i; x=i/5,y=i%5; for(i=0;i<8;i++) { int newx=x+dx[i]; int newy=y+dy[i]; if(newx>=0&&newx<5&&newy>=0&&newy<5) { state &u=st[rear]; memcpy(&u,&p,sizeof(p)); u[x*5+y]=u[newx*5+newy]; u[newx*5+newy]=2; dist[rear]=dist[front]+1; if(try_to_insert(rear)) rear++; } } front++; } } int main() { cin>>s; string str; getline(cin,str); while(s--) { memset(dist,0,sizeof(dist)); int i,j; for(i=0;i<5;i++) { getline(cin,str); for(j=0;j<5;j++) { if(str[j]!=' ') st[1][i*5+j]=str[j]-'0'; else st[1][i*5+j]=2; } } int d=bfs(); if(d<=0) cout<<"Unsolvable in less than 11 move(s)."<