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
First line of the input file contains an integer N (N<14 are="are" below:="below:" description="description" each="each" given="given" how="how" indicates="indicates" inputs="inputs" is="is" many="many" of="of" p="p" set="set" sets="sets" that="that" the="the" there.="there.">
Sample Input
Sample Output
Each set consists of five lines; each line represents one row of a chessboard. The positions occupied by white knights are marked by 0 and the positions occupied by black knights are marked by 1. The space corresponds to the empty square on board.
There is no blank line between the two sets of input.
The first set of the sample input below corresponds to this configuration:
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.”
DFS+ bitmask + backtracking
#include<stdio.h>
#include<cstring>
#include<string>
#include<vector>
#include<set>
#include<queue>
#define REP(i, b, n) for (int i = b; i < n; i++)
#define rep(i, n) REP(i, 0, n)
#define DBG 0
using namespace std;
struct Node{
Node(){}
Node(int a,int b,int c):val(a),row(b),col(c){}
int val,row,col;
bool operator <(Node b)const{return val<b.val||row<b.row||col<b.col;}
};
int T,bnum=13;
set<Node>st;
int bottom[]= {5,10,11,12,15,16,17,18,20,21,22,23,24};
char chess[5][10],dir[8][2];
void init(){
rep(i,5)T<<=1,T+=1;
T<<=1;
rep(i,4)T<<=1,T+=1;
rep(i,3)T<<=1; rep(i,2)T<<=1,T+=1;
rep(i,4)T<<=1;T<<=1,T+=1;
rep(i,5)T<<=1;
dir[0][0]=-2,dir[0][1]=-1,
dir[1][0]=-2,dir[1][1]=1,
dir[2][0]=-1,dir[2][1]=-2,
dir[3][0]=-1,dir[3][1]=2,
dir[4][0]=1,dir[4][1]=-2,
dir[5][0]=1,dir[5][1]=2,
dir[6][0]=2,dir[6][1]=-1,
dir[7][0]=2,dir[7][1]=1;
}
void getHole(int &a,int &b){
rep(i,5)rep(j,5)if(chess[i][j]==' ')a=i,b=j;
}
bool getH(int a,int b){
return (a>>24-b)%2;
}
int setH(int a,int pos,int b){
if(b)a|=1<<(24-pos);
else a-=(a & (1<<(24-pos)) );
return a;
}
int count(int c){
int a=0;
rep(i,bnum){if(getH(c,bottom[i]))a++;}
if(DBG)printf("cnt: %d\n",a);
return a;
}
int getC(){
int c=0;
rep(i,5)rep(j,5){
if(chess[i][j]=='1')c<<=1,c+=1;
else c<<=1;
}
return c;
}
int move(int v,int cr,int cc,int nr,int nc){
int from=cr*5+cc,to=nr*5+nc;
int h=getH(v,from);
v=setH(v,from,0);
v=setH(v,to,h); //move
return v;
}
bool valid(int a,int b){return a>=0&&a<5&&b>=0&&b<5;}
void print(int a){
rep(i,5){
rep(j,5)printf("%d",getH(a,i*5+j));
printf("\n");
}
}
void ans(){
int cur,dis,next,cr,cc,nr,nc;
bool h;
Node nd;
int res=11;
st.clear();
getHole(cr,cc);
queue<pair<Node,int> >q;
nd=Node(getC(),cr,cc);
st.insert(nd);
q.push(make_pair(nd,0));
while(q.size()){
cur=q.front().first.val,dis=q.front().second,
cr=q.front().first.row,cc=q.front().first.col;
if(DBG){printf("cur\n");print(cur);printf("cur cr cc dis %d %d %d %d\n",cur,cr,cc,dis);}
q.pop();
if(dis>10)break;
if(count(cur)>10-dis)continue;
if(cur==T&&cr==2&&cc==2){res=dis;break;}
rep(i,8){
nr=cr+dir[i][0],nc=cc+dir[i][1];
if(valid(nr,nc)){
if(DBG)printf("new hole %d %d\n",nr,nc);
next=move(cur,nr,nc,cr,cc);
nd=Node(next,nr,nc);
if(!st.count(nd))
st.insert(nd),q.push(make_pair(Node(next,nr,nc),dis+1));
}
}
}
if(res>10)printf("Unsolvable in less than 11 move(s).\n");
else printf("Solvable in %d move(s).\n",res);
}
int main(){
int n;
scanf("%d ",&n);
init();
rep(i,n){
rep(i,5)fgets(chess[i],10,stdin);
ans();
}
}
(1<=i<=5) that will yield the value 23.
: {1,2,3,4,5} -> {1,2,3,4,5} is a bijective function
{+,-,*} (1<=i<=4)
) and the number of edges m (
). The next m lines describe the m edges. Each edge is given by the numbers of the two nodes connected by it. Nodes are numbered from 0 to n-1. Edges are undirected. Nodes have degrees of three or less. The network is not neccessarily connected.
into each circle separately, and the sum of numbers in two adjacent circles should be a prime.
. The next l lines of input will be the matrix itself, each line will contain l uppercase letters.
. Each solution will consist of 8 numbers. Each of the 8 numbers will be the ROW coordinate for that solution. The column coordinate will be indicated by the order in which the 8 numbers are printed. That is, the first number represents the ROW in which the queen is positioned in column 1; the second number represents the ROW in which the queen is positioned in column 2, and so on.
8 representation of each solution is shown below.
). Then, you are given a list of pairs of places p and q, one pair per line, with the places on each line separated by white-space. The pair ``
" indicates that César can visit q after p.