1675: [Usaco2005 Feb]Rigging the Bovine Election 竞选划区
Time Limit: 5 Sec Memory Limit: 64 MB
Submit: 196 Solved: 116
[Submit][Status][Discuss]
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
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
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.