马的遍历

题目描述

有一个n*m的棋盘(1 < n,m<=400),在某个点上有一个马,要求你计算出马到达棋盘上任意一个点最少要走几步

输入输出格式

输入格式:
一行四个数据,棋盘的大小和马的坐标

输出格式:
一个n*m的矩阵,代表马到达某个点最少要走几步(左对齐,宽5格,不能到达则输出-1)

输入输出样例

输入样例#1:
3 3 1 1

输出样例#1:
0 3 2
3 -1 1
2 1 4
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分析:

这是一题十分耀眼的广搜!!!
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程序:
const
dx:array[1..8] of longint=(2,1,-1,-2,-2,-1,2,1);
dy:array[1..8] of longint=(-1,-2,-2,-1,1,2,1,2);
var
s:array[1..233,1..233] of longint;
h:array[1..233333,1..2] of longint;
bz:array[1..233,1..233] of boolean;
n,m,x,y,i,j,p:longint;
procedure bfs(c,d:longint);
var
i,j,t,w,x,y:longint;
begin
    bz[c,d]:=false;
    t:=1; w:=1;
    h[t,1]:=c; h[t,2]:=d;
    s[c,d]:=1;
    repeat
         for i:=1 to 8 do
         begin
             x:=h[t,1]+dx[i];
             y:=h[t,2]+dy[i];
             if (x>0)and(x<=n)and(y>0)and(y<=m)and bz[x,y] then
             begin
                 inc(w);
                 h[w,1]:=x;
                 h[w,2]:=y;
                 bz[x,y]:=false;
                 s[x,y]:=s[h[t,1],h[t,2]]+1;
             end;
         end;
         inc(t);
    until t>w;
end;
begin
    readln(n,m,x,y);
    fillchar(s,sizeof(s),0);
    fillchar(bz,sizeof(bz),true);
    bfs(x,y);
    for i:=1 to n do
    begin                                    
        for j:=1 to m do
        begin
            write(s[i,j]-1);
            if (s[i,j]-1<10)and(s[i,j]-1>=0)then write('    ');
            if ((s[i,j]-1<100)and(s[i,j]-1>=10))or(s[i,j]-1=-1) then write('   ');
            if (s[i,j]-1<1000)and(s[i,j]-1>=100) then write('  ');
            if s[i,j]-1>=1000 then write(' ');
        end;
        writeln;
    end;
end.
原文地址:https://www.cnblogs.com/YYC-0304/p/9499994.html