【BZOJ】1675: [Usaco2005 Feb]Rigging the Bovine Election 竞选划区（暴力dfs+set判重）

http://www.lydsy.com/JudgeOnline/problem.php?id=1675

一开始我写了个枚举7个点。。。。。。。

但是貌似。。。

写挫了。

然后我就写dfs。。

判重好难写啊。

。。。。

本来用hash的。。

但是对拍一直wa。。

所以干脆用set。。

然后将数值调大。。

然后就过了。。

然后bzoj数据弱。。

自己对拍还是hash有冲突的。。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
#include <set>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << #x << " = " << x << endl
#define printarr(a, n, m) for1(aaa, 1, n) { for1(bbb, 1, m) cout << a[aaa][bbb]; cout << endl; }
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; }

const int md=1000000007;
int mp[7][7], vis[7][7], ans;
set<int> hs;
const int dx[]={-1, 1, 0, 0}, dy[]={0, 0, -1, 1};
int Hash() {
	int ret=0, k=7;
	for1(i, 1, 5) for1(j, 1, 5) if(vis[i][j]) {
		ret=(ret+(((mp[i][j]*k)%md)*(i*5+j)%md))%md;
		k=(k*7)%md;
	}
	return ret;
}
bool check(int x, int y) {
	rep(i, 4) if(vis[x+dx[i]][y+dy[i]]) return 1;
	return 0;
}
void dfs(int now, int a, int b) {
	if(now==8) {
		if(a>b) {
			int h=Hash();
			if(hs.count(h)==0) { hs.insert(h); ++ans; }
		}
		return;
	}
	for1(i, 1, 5) for1(j, 1, 5) if(!vis[i][j] && (check(i, j) || now==1)) {
		vis[i][j]=1;
		dfs(now+1, a+(mp[i][j]==31), b+(mp[i][j]==43));
		vis[i][j]=0;
	}
}

int main() {
	for1(i, 1, 5) for1(j, 1, 5) {
		char ch; for(ch=getchar(); !(ch=='H'||ch=='J'); ch=getchar());
		if(ch=='J') mp[i][j]=31;
		else mp[i][j]=43;
	}
	dfs(1, 0, 0);
	print(ans);
	return 0;
}

后边写的枚举每个点

#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << #x << " = " << x << endl
#define printarr(a, n, m) for1(aaa, 1, n) { for1(bbb, 1, m) cout << a[aaa][bbb]; cout << endl; }
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; }

int mp[7][7], ans, xy[10], x[10], y[10], vis[7][7], vv[7][7], cnt;
const int dx[]={-1, 1, 0, 0}, dy[]={0, 0, -1, 1};
void dfs(int x, int y) {
	vv[x][y]=1;
	++cnt;
	rep(i, 4) {
		int fx=dx[i]+x, fy=dy[i]+y;
		if(vis[fx][fy] && !vv[fx][fy]) dfs(fx, fy);
	}
}
bool check() {
	for1(i, 1, 7) x[i]=(xy[i]+4)/5, y[i]=xy[i]-(x[i]-1)*5;
	CC(vis, 0); CC(vv, 0);
	for1(i, 1, 7) vis[x[i]][y[i]]=1;
	cnt=0;
	dfs(x[1], y[1]);
	if(cnt!=7) return 0;
	int sum=0;
	for1(i, 1, 7) sum+=mp[x[i]][y[i]];
	return sum>3;
}
int main() {
	for1(i, 1, 5) for1(j, 1, 5) {
		char ch; for(ch=getchar(); ch!='H'&&ch!='J'; ch=getchar());
		mp[i][j]=ch=='J';
	}
	for(xy[1]=1; xy[1]<=19; ++xy[1])
		for(xy[2]=xy[1]+1; xy[2]<=20; ++xy[2])
			for(xy[3]=xy[2]+1; xy[3]<=21; ++xy[3])
				for(xy[4]=xy[3]+1; xy[4]<=22; ++xy[4])
					for(xy[5]=xy[4]+1; xy[5]<=23; ++xy[5])
						for(xy[6]=xy[5]+1; xy[6]<=24; ++xy[6])
							for(xy[7]=xy[6]+1; xy[7]<=25; ++xy[7])
								if(check()) ++ans;
	print(ans);
	return 0;
}

 

 

---

 

 

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

Silver