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

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( Oleft( {25}^{7} ight) )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）

 1 /**************************************************************
 2     Problem: 1675
 3     User: HansBug
 4     Language: Pascal
 5     Result: Accepted
 6     Time:1860 ms
 7     Memory:228 kb
 8 ****************************************************************/
 9  
10 const
11      mx:array[1..4] of longint=(0,1,0,-1);
12      my:array[1..4] of longint=(1,0,-1,0);
13 var
14    i,j,k,l,m,n,ans,i1,i2,i3,i4,i5,i6,i7:longint;
15    ch:char;
16    map:array[0..6,0..6] of longint;
17    s:array[0..8] of longint;
18 function check:boolean;
19          var
20             i,j,sx,sy,xx,yy,sum,t,w:longint;
21             nx,ny:array[0..8] of longint;
22             q:array[0..10] of longint;
23             d:array[0..6,0..6] of longint;
24          begin
25               sum:=0;t:=0;w:=1;
26               fillchar(nx,sizeof(nx),0);
27               fillchar(ny,sizeof(ny),0);
28               fillchar(q,sizeof(q),0);
29               fillchar(d,sizeof(d),0);
30               for i:=1 to 7 do
31                   begin
32                        ny[i]:=s[i] mod 5;
33                        nx[i]:=(s[i] div 5)+1;
34                        if ny[i]=0 then
35                           begin
36                                ny[i]:=5;
37                                dec(nx[i]);
38                           end;
39                        d[nx[i],ny[i]]:=i;
40                   end;
41               q[1]:=1;d[nx[1],ny[1]]:=0;sum:=map[nx[1],ny[1]];
42               while t<w do
43                     begin
44                          inc(t);
45                          xx:=nx[q[t]];
46                          yy:=ny[q[t]];
47                          for k:=1 to 4 do
48                              begin
49                                   sx:=xx+mx[k];
50                                   sy:=yy+my[k];
51                                   if (sx<1) or (sy<1) or (sy>5) or (sy>5) or (d[sx,sy]=0) then continue;
52                                   inc(w);
53                                   q[w]:=d[sx,sy];
54                                   d[sx,sy]:=0;
55                                   inc(sum,map[sx,sy]);
56                              end;
57                     end;
58               exit((w=7) and (sum>3));
59          end;
60 begin
61      fillchar(map,sizeof(map),0);
62      for i:=1 to 5 do
63          for j:=1 to 5 do
64              begin
65                   read(ch);
66                   if ch='J' then map[i,j]:=1 else map[i,j]:=0;
67                   if j=5 then readln;
68              end;
69      ans:=0;
70      for i1:=1 to 19 do
71          for i2:=i1+1 to 20 do
72              for i3:=i2+1 to 21 do
73                  for i4:=i3+1 to 22 do
74                      for i5:=i4+1 to 23 do
75                          for i6:=i5+1 to 24 do
76                              for i7:=i6+1 to 25 do
77                                  begin
78                                       s[1]:=i1;s[2]:=i2;s[3]:=i3;
79                                       s[4]:=i4;s[5]:=i5;s[6]:=i6;s[7]:=i7;
80                                       if check then inc(ans);
81                                  end;
82      writeln(ans);
83      readln;
84 end.