bzoj 1675: [Usaco2005 Feb]Rigging the Bovine Election 竞选划区（暴力）

1675: [Usaco2005 Feb]Rigging the Bovine Election 竞选划区

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 285  Solved: 192

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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

C(25, 7) =
480700

放心的暴力吧

#include<stdio.h>
#include<string.h>
char str[7][7];
int ans, temp, c[8], d[8], flag[8];
int dir[4][2] = {1,0,0,1,-1,0,0,-1};
void Jud(int p)
{
int i, j, dx, dy;
flag[p] = 1, temp++;
for(i=0;i<=3;i++)
{
dx = c[p]+dir[i][0];
dy = d[p]+dir[i][1];
for(j=1;j<=7;j++)
{
if(c[j]==dx && d[j]==dy && flag[j]==0)
Jud(j);
}
}
}
void Sech(int x, int y)
{
int i, sum;
if(y==8)
{
temp = sum = 0;
memset(flag, 0, sizeof(flag));
for(i=1;i<=7;i++)
{
if(str[c[i]][d[i]]=='J')
sum++;
}
Jud(1);
if(sum>=4 && temp==7)
ans++;
return;
}
if(x>25)
return;
for(i=x;i<=25;i++)
{
c[y] = (i+4)/5;
d[y] = i-((i+4)/5-1)*5;
Sech(i+1, y+1);
}
}
int main(void)
{
int i;
for(i=1;i<=5;i++)
scanf("%s", str[i]+1);
Sech(1, 1);
printf("%d\n", ans);
return 0;
}