1675: [Usaco2005 Feb]Rigging the Bovine Election 竞选划区（题解第一弹）

1675: [Usaco2005 Feb]Rigging the Bovine Election 竞选划区

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 196  Solved: 116
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Description

It's election time. The farm is partitioned into a 5x5 grid of cow locations, each of which holds either a Holstein ('H') or Jersey ('J') cow. The Jerseys want to create a voting district of 7 contiguous (vertically or horizontally) cow locations such that the Jerseys outnumber the Holsteins. How many ways can this be done for the supplied grid?

农场被划分为5x5的格子，每个格子中都有一头奶牛，并且只有荷斯坦(标记为H)和杰尔西（标记为J）两个品种．如果一头奶牛在另一头上下左右四个格子中的任一格里，我们说它们相连．    奶牛要大选了．现在有一只杰尔西奶牛们想选择7头相连的奶牛，划成一个竞选区，使得其中它们品种的奶牛比荷斯坦的多．  要求你编写一个程序求出方案总数．

Input

* Lines 1..5: Each of the five lines contains five characters per line, each 'H' or 'J'. No spaces are present.

    5行，输入农场的情况．

Output

* Line 1: The number of distinct districts of 7 connected cows such that the Jerseys outnumber the Holsteins in the district.

    输出划区方案总数．

Sample Input

HHHHH
JHJHJ
HHHHH
HJHHJ
HHHHH

Sample Output

2

HINT

Source

Silver

题解：一开始这道题我看了半天硬是没敢做，首先想到的是状压DP，可是状压虽然25格应该能压得下，但是怎么判断连续？所以一时束手无策

直到看到了某位网上的神犇说——大力出奇迹！！！orzorzorz

然后我就按照那样子来了一发，结果居然能A！！！神奇QAQ

实际上就是\( O\left( {25}^{7} \right) \)的朴素暴力，实在没想到这都能AC，于是有种瞬间被彻底吓尿的感觉QAQ

但是很明显有很大优化的空间，于是本人打算明天再来一发强化版的么么哒（现已完成，传送门）

 

 /**************************************************************
     Problem:
     User: HansBug
     Language: Pascal
     Result: Accepted
     Time: ms
     Memory: kb
 ****************************************************************/

 const
      mx:array[..] of longint=(,,,-);
      my:array[..] of longint=(,,-,);
 var
    i,j,k,l,m,n,ans,i1,i2,i3,i4,i5,i6,i7:longint;
    ch:char;
    map:array[..,..] of longint;
    s:array[..] of longint;
 function check:boolean;
          var
             i,j,sx,sy,xx,yy,sum,t,w:longint;
             nx,ny:array[..] of longint;
             q:array[..] of longint;
             d:array[..,..] of longint;
          begin
               sum:=;t:=;w:=;
               fillchar(nx,sizeof(nx),);
               fillchar(ny,sizeof(ny),);
               fillchar(q,sizeof(q),);
               fillchar(d,sizeof(d),);
               for i:= to  do
                   begin
                        ny[i]:=s[i] mod ;
                        nx[i]:=(s[i] div )+;
                        if ny[i]= then
                           begin
                                ny[i]:=;
                                dec(nx[i]);
                           end;
                        d[nx[i],ny[i]]:=i;
                   end;
               q[]:=;d[nx[],ny[]]:=;sum:=map[nx[],ny[]];
               while t<w do
                     begin
                          inc(t);
                          xx:=nx[q[t]];
                          yy:=ny[q[t]];
                          for k:= to  do
                              begin
                                   sx:=xx+mx[k];
                                   sy:=yy+my[k];
                                   if (sx<) or (sy<) or (sy>) or (sy>) or (d[sx,sy]=) then continue;
                                   inc(w);
                                   q[w]:=d[sx,sy];
                                   d[sx,sy]:=;
                                   inc(sum,map[sx,sy]);
                              end;
                     end;
               exit((w=) and (sum>));
          end;
 begin
      fillchar(map,sizeof(map),);
      for i:= to  do
          for j:= to  do
              begin
                   read(ch);
                   if ch='J' then map[i,j]:= else map[i,j]:=;
                   if j= then readln;
              end;
      ans:=;
      for i1:= to  do
          for i2:=i1+ to  do
              for i3:=i2+ to  do
                  for i4:=i3+ to  do
                      for i5:=i4+ to  do
                          for i6:=i5+ to  do
                              for i7:=i6+ to  do
                                  begin
                                       s[]:=i1;s[]:=i2;s[]:=i3;
                                       s[]:=i4;s[]:=i5;s[]:=i6;s[]:=i7;
                                       if check then inc(ans);
                                  end;
      writeln(ans);
      readln;
 end.