Section 2.1 The Castle

最新推荐文章于 2025-08-15 10:51:23 发布

YY19970227

最新推荐文章于 2025-08-15 10:51:23 发布

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分类专栏： USACO 2 文章标签： ansj output pascal input css

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本文介绍了一种使用深度优先搜索(DFS)解决迷宫问题的方法，并通过FloodFill算法来寻找最大连通区域，同时考虑了墙壁拆分对迷宫路径的影响。该算法能够找出拆除特定墙壁后形成的两个相邻区域的合并所能达到的最大区域数量。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

先用DFS做一个Flood Fill

然后枚举墙并判断记住按照题目要求枚举

具体看标程这题是挺难的

{
ID: yaoyuan4
PROG: castle
LANG: PASCAL
}
Program castle;
type
css = record
   west, north, east, south : boolean;
end;
const
fx : array[1..4] of longint = (1, 2, 4, 8);
inf = 'castle.in'; outf = 'castle.out';
var
m, n, ans, ansi, ansj, amx : longint;
f : array[0..100, 0..100] of css;
opi : array[0..100, 0..100] of longint;
dj : array[0..10000] of longint;
ansk : char;
Procedure vaei(x : longint; var y : css);
var
   i, k : longint;
begin
   for i := 4 downto 1 do
    if x >= fx[i] then
     begin
      dec(x, fx[i]);
      with y do
       case i of
        4 : south := true;
        3 : east := true;
        2 : north := true;
        1 : west := true;
       end;
     end;
end;
Procedure init;
var
   i, j, k : longint;
begin
   assign(input, inf); reset(input);
   readln(m, n);
   for i := 1 to n do
    begin
     for j := 1 to m do
      begin
       read(k);
       vaei(k, f[i, j]);
      end;
     readln;
    end;
   close(input);
   fillchar(opi, sizeof(opi), 0);
   fillchar(dj, sizeof(dj), 0);
   ans := 0; amx := 0;
end;
Procedure dfs(x, y : longint);
var
   i : longint;
begin
   opi[x, y] := dj[0];
   inc(dj[dj[0]]);
   if y > 1 then
    if (not(f[x, y].west)) and (opi[x, y-1] = 0) then dfs(x, y-1);
   if x > 1 then
    if (not(f[x, y].north)) and (opi[x-1, y] = 0) then dfs(x-1, y);
   if y < m then
    if (not(f[x, y].east)) and (opi[x, y+1] = 0) then dfs(x, y+1);
   if x < n then
    if (not(f[x, y].south)) and (opi[x+1, y] = 0) then dfs(x+1, y);
end;
Procedure work;
var
   i, j, ts : longint;
begin
   for i := 1 to n do
    for j := 1 to m do
     if opi[i, j] = 0 then
      begin
       inc(dj[0]);
       dfs(i, j);
      end;
    for i := 1 to dj[0] do
     if dj[i] > amx then amx := dj[i];
    for j := 1 to m do
     for i := n downto 1 do
      begin
       if (f[i, j].north) and (i > 1) and (opi[i, j] <> opi[i-1, j]) then
        begin
         ts := dj[opi[i, j]] + dj[opi[i-1, j]];
         if ts > ans then
          begin
           ans := ts;
           ansi := i;
           ansj := j;
           ansk := 'N';
          end;
        end;
       if (f[i, j].east) and (j < m) and (opi[i, j] <> opi[i, j+1]) then
        begin
         ts := dj[opi[i, j]] + dj[opi[i, j+1]];
         if ts > ans then
          begin
           ans := ts;
           ansi := i;
           ansj := j;
           ansk := 'E';
          end;
        end;
      end;
end;
Procedure print;
var
   i : longint;
begin
   assign(output, outf); rewrite(output);
   writeln(dj[0]);
   writeln(amx);
   writeln(ans);
   writeln(ansi,' ',ansj,' ',ansk);
   close(output);
end;
begin
init;
work;
print;
end.