bzoj1675 [Usaco2005 Feb]Rigging the Bovine Election 竞选划区

 

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

 

 

暴力出奇迹啊……

直接7个for枚举位置判可行性也能过啊……

我服了

#include<cstdio>
const int mx[4]={0,1,0,-1};
const int my[4]={1,0,-1,0};
bool map[6][6];
int d[6][6];
int s[8];
int nx[8],ny[8];
int q[10];
int ans;
inline bool mark()
{
	int i,j,sx,sy,xx,yy,sum=0,t=0,w=1;
	for (i=1;i<=5;i++)for(j=1;j<=5;j++)d[i][j]=0;
	for (i=1;i<=7;i++)
	{
		ny[i]=s[i]%5;nx[i]=s[i]/5;
		if (ny[i])nx[i]++;
		if (!ny[i])ny[i]=5;
		d[nx[i]][ny[i]]=i;
	}
	q[1]=1;d[nx[1]][ny[1]]=0;sum=map[nx[1]][ny[1]];
	while (t<w)
	{
		xx=nx[q[++t]];
		yy=ny[q[t]];
		for (int k=0;k<4;k++)
		{
		  sx=xx+mx[k];sy=yy+my[k];
		  if (sx>0&&sy>0&&sx<6&&sy<6&&d[sx][sy])
		  {
		  	q[++w]=d[sx][sy];
		  	d[sx][sy]=0;
		  	sum+=map[sx][sy];
		  }
		}
	}
	return w==7&&sum>3;
}
int main()
{
	for (int i=1;i<=5;i++)
	for (int j=1;j<=5;j++)
	{
		char ch=getchar();
		while (ch!='H'&&ch!='J')ch=getchar();
		if (ch=='J')map[i][j]=1;
	}
	for (s[1]=1;s[1]<=19;s[1]++)
	for (s[2]=s[1]+1;s[2]<=20;s[2]++)
	for (s[3]=s[2]+1;s[3]<=21;s[3]++)
	for (s[4]=s[3]+1;s[4]<=22;s[4]++)
	for (s[5]=s[4]+1;s[5]<=23;s[5]++)
	for (s[6]=s[5]+1;s[6]<=24;s[6]++)
	for (s[7]=s[6]+1;s[7]<=25;s[7]++)
	  if (mark())ans++;
	printf("%d",ans);
}