Pascal Knight Travel, a program for pascal Knight Travel

If you want, leave an email and I'll send it to you.

Dynamic programming. Dynamic equations: f[,j-2] or f[i-2,j-1] (Note that it's a Boolean!) )

Although the input is different.

But you can change it, the ** of the title.

constmaxm=50;

maxn=50;

varm,n,x1,y1,x2,y2:integer;

i,j,k,x,y:integer;

map:array[-2..maxm+2,-2..maxn+2] of extended;

beginfillchar(map,sizeof(map),0);

readln(m,n,x1,y1,x2,y2);

map[x1,y1]:=1;

for i:=x1+1 to x2 do

for j:=1 to m do

map[i,j]:=map[i-1,j-2]+map[i-1,j+2]+map[i-2,j-1]+map[i-2,j+1];

writeln(map[x2,y2]:0:0);

end.

Knight's Journey (Pascal).

There is a chessboard with n*m (2<=n<=50, 2<=m<=50), and there is a Chinese chess horse at any point on the board, and the rules for the horse to move are: 1Horse Walking Japanese Character 2Horses can only go to the right.

Task: When n and m are given, give the position of the horse's starting point and the position of the end point at the same time, and try to find the number of all paths from the beginning to the end point. For example: (n=10, m=10), (1,5) (start), (3,5) (end).

Output: 2 (i.e. 2 paths from (1,5) to (3,5)).

Input format: n, m, x1, y1, x2, y2 (represent n, m, start coordinates, end coordinates, respectively).

Output format: Number of paths (0 if no path from start to end does not exist).

From the (x1,y1) position, define the direction of the stages in order from left to right. The set of vault positions to the left of (x2,y2) up to (x2,y2) are all subproblems of (x2,y2), and the number of paths from the starting point to (x2,y2) is actually equal to the sum of the number of paths from the starting point to these sets of positions. You can calculate the number of paths at each point in each stage in the order of the phases.

Phase I: The current column position of the Chinese chess horse (x1 i x2).

State J: The position of the Chinese chess horse in the row of column i (1 i m).

The state transition equation map[i,j]: the number of paths from the starting point (x1,y1) to (i,j).

The specific algorithm is as follows:

fillchar(map,sizeof(map),0);

map[x1,y1]←1；

for i←x1+1 to x2 do

for j←1 to m do

map[i,j]←map[i-1,j-2]+map[i-1,j+2]+map[i-2,j-1]+map[i-2,j+1];

writeln(map[x2,y2]);

Reference program】program qishiyouli;

constmaxm=50;

maxn=50;

varm,n,x1,y1,x2,y2:integer;

i,j,k,x,y:integer;

map:array[-2..maxm+2,-2..maxn+2] of extended;

beginfillchar(map,sizeof(map),0);

readln(m,n,x1,y1,x2,y2);

map[x1,y1]:=1;

for i:=x1+1 to x2 do

for j:=1 to m do

map[i,j]:=map[i-1,j-2]+map[i-1,j+2]+map[i-2,j-1]+map[i-2,j+1];

writeln(map[x2,y2]:0:0);

end.

Rub the experience.

Rub the experience. Rub the experience.

Rub the experience. Rub the experience.

If you ask for all the solutions, you have to test 4 36= kinds, and the computer is too busy.

The description of the title is not detailed enough.

The problem is in terms of output.

If the coordinates of the upper left corner of the board are (1,1) and the coordinates of the lower right corner are (n,m), then dfs(0,0,0); to change to dfs(1,1,0);

writeln('(xy[j,1],'xy[j,2],'Changed to writeln('(xy[j,1],'m-xy[j,2]+1,')

writeln('(n,' m,') to writeln('(n,' 1,')

Question: Chessboard, n*m, Requirements:

The rules for horses to walk are:1Horse Walking Japanese Character 2Horses can only go to the right.

When n and m are entered, find a path from the bottom left corner to the top right corner.

Re: Actually, your program is fine, after many debuggings, it has always been correct. In our fixed thinking, the first square on the board is (0,0), but if there is a special rule for the problem, then follow the rules. It's the same with you.

The above is my reply, please decide.

This is a classic type of backtracking, and there are many solutions on the Internet, you can check it out.

It's all possible, but there is one thing you don't change when you modify the program.

program aaa;

constn=8; nsq=n*n;

typeindex=1..n;

vari,j:index;

q:boolean;

a:array[1..2,1..8]of integer;

b:array[1..n,1..n]of integer;

procedure try(x,y:index;i:integer;var q:boolean);

vark,u,v:integer;

q1:boolean;

begink:=0;

repeat

k:=k+1;

q1:=false;

u:=x+a[1,k];

v:=y+a[2,k];

if (u>=1) and (u<=n) and(v>=1) and(v<=n) then

if b[u,v]=0 then

beginb[u,v]:=i;

if ibegin

try(u,v,i+1,q1);

if not q1 then b[u,v]:=0;

endelse q1:=true;

end;until q1 or(k=8);

q:=q1;

end;begin

a[1,1]:=-1;a[1,2]:=-2;a[1,3]:

2;a[1,4]:=-1;a[1,5]:=1;a[1,6]:

2;a[1,7]:=2;a[1,8]:=1;

a[2,1]:=2;a[2,2]:=1;a[2,3]:

1;a[2,4]:=-2;a[2,5]:=-2;a[2,6]:

1;a[2,7]:=1;a[2,8]:=2;

for i:=1 to n do

for j:=1 to n do

b[i,j]:=0;

b[1,1]:=1;

try(1,1,2,q);

if q then

for i:=1 to n do

beginfor j:=1 to n do

write(b[i,j]:5);

writeln;

endelse writeln('no solution!');

end.

if (x[j,1]+a[i,1]< n) and (x[j,2]+a[i,2]< m) and (x[j,1]+a[i,1]>=0) and (x[j,2]+a[i,2]>=0) then

It should be if (x[j,1]+a[i,1]<=n) and (x[j,2]+a[i,2]<=m) and (x[j,1]+a[i,1]>=0) and (x[j,2]+a[i,2]>=0) then

If it is not equal then it will not be reached.

Knight travels have always been judged by RP.

I might as well make it up again, I can do it, the first time it's right, the second time it's wrong, and the third time it's right...

1. Conditional judgment 2. Array burst stack.