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: 1675
    User: HansBug
    Language: Pascal
    Result: Accepted
    Time:1860 ms
    Memory:228 kb
****************************************************************/
 
const
     mx:array[1..4] of longint=(0,1,0,-1);
     my:array[1..4] of longint=(1,0,-1,0);
var
   i,j,k,l,m,n,ans,i1,i2,i3,i4,i5,i6,i7:longint;
   ch:char;
   map:array[0..6,0..6] of longint;
   s:array[0..8] of longint;
function check:boolean;
         var
            i,j,sx,sy,xx,yy,sum,t,w:longint;
            nx,ny:array[0..8] of longint;
            q:array[0..10] of longint;
            d:array[0..6,0..6] of longint;
         begin
              sum:=0;t:=0;w:=1;
              fillchar(nx,sizeof(nx),0);
              fillchar(ny,sizeof(ny),0);
              fillchar(q,sizeof(q),0);
              fillchar(d,sizeof(d),0);
              for i:=1 to 7 do
                  begin
                       ny[i]:=s[i] mod 5;
                       nx[i]:=(s[i] div 5)+1;
                       if ny[i]=0 then
                          begin
                               ny[i]:=5;
                               dec(nx[i]);
                          end;
                       d[nx[i],ny[i]]:=i;
                  end;
              q[1]:=1;d[nx[1],ny[1]]:=0;sum:=map[nx[1],ny[1]];
              while t<w do
                    begin
                         inc(t);
                         xx:=nx[q[t]];
                         yy:=ny[q[t]];
                         for k:=1 to 4 do
                             begin
                                  sx:=xx+mx[k];
                                  sy:=yy+my[k];
                                  if (sx<1) or (sy<1) or (sy>5) or (sy>5) or (d[sx,sy]=0) then continue;
                                  inc(w);
                                  q[w]:=d[sx,sy];
                                  d[sx,sy]:=0;
                                  inc(sum,map[sx,sy]);
                             end;
                    end;
              exit((w=7) and (sum>3));
         end;
begin
     fillchar(map,sizeof(map),0);
     for i:=1 to 5 do
         for j:=1 to 5 do
             begin
                  read(ch);
                  if ch='J' then map[i,j]:=1 else map[i,j]:=0;
                  if j=5 then readln;
             end;
     ans:=0;
     for i1:=1 to 19 do
         for i2:=i1+1 to 20 do
             for i3:=i2+1 to 21 do
                 for i4:=i3+1 to 22 do
                     for i5:=i4+1 to 23 do
                         for i6:=i5+1 to 24 do
                             for i7:=i6+1 to 25 do
                                 begin
                                      s[1]:=i1;s[2]:=i2;s[3]:=i3;
                                      s[4]:=i4;s[5]:=i5;s[6]:=i6;s[7]:=i7;
                                      if check then inc(ans);
                                 end;
     writeln(ans);
     readln;
end.