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
usaco良心网站，直接暴力不会挂……233
暴力枚举点就行了，每次枚举就在已经选到的点周围选就行咯，随手带几个剪枝。
至于去重，直接hash就好。在累计答案处hash：64MS，每一层搜索都hash（去掉重复扩展的状态）20MS，玩火人工二分MOD:44MS、48MS……
运气#1
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MOD=10001;
int ans=0,xx,yy,k;
int hash[MOD];
char c[5][5];
void dfs(int x,int y,int hn,int jn,int p){
if (x<0||y<0||x>4||y>4) return;
if (x<xx||(x==xx&&y<yy)) return;
if (c[x][y]==‘H‘) hn++;else jn++;
if (hn>3) return;
p+=1<<((x*5+y));
k=p%MOD;
while(hash[k]!=-1){
if (hash[k]==p) return;
k++;
if (k>=MOD) k-=MOD;
}
hash[k]=p;
if (hn+jn==7){
ans++;
return;
}
char s=c[x][y];
c[x][y]=0;
for (int i=0;i<5;i++)
for (int j=0;j<5;j++)
if (c[i][j]==0){
if (c[i+1][j]!=0) dfs(i+1,j,hn,jn,p);
if (c[i-1][j]!=0) dfs(i-1,j,hn,jn,p);
if (c[i][j+1]!=0) dfs(i,j+1,hn,jn,p);
if (c[i][j-1]!=0) dfs(i,j-1,hn,jn,p);
}
c[x][y]=s;
}
int main(){
for (int i=0;i<5;i++) scanf("%s",c[i]);
memset(hash,-1,sizeof(hash));
for (xx=0;xx<5;xx++)
for (yy=0;yy<5;yy++) dfs(xx,yy,0,0,0);
printf("%d\n",ans);
}