Problem 3 银河之星(galaxy.cpp/c/pas)

数据组数不超过10.

这题就是记忆化搜索

9点染色减少状态，map记忆化

b[i][j][k]表示棋子可否从k方向到(i,j)

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<map>
using namespace std;
#define MAXN (100+10)
#define MAXK (10+10)
#define MAXDIV (4)
int k,n,m,x2,y2,a[MAXDIV][MAXDIV],b[MAXDIV][MAXDIV][9];
int c[8][2]={{0,1},{0,2},{1,0},{2,0},{1,1},{2,2},{1,2},{2,1}};  //↑↓→←↗↙↘↖
map<long long,int> h;
bool is_ok()
{
	int ans=0;
	for (int i=0;i<3;i++)
		for (int j=0;j<3;j++)
			ans=(11*ans+a[i][j]);
	if (h.find(ans)!=h.end()) return 0;
	return h[ans]=1;
}
bool dfs(int x)
{
	/*
	for (int j=3;j;j--)
		{
			for (int i=1;i<=3;i++)
				cout<<a[i%3][j%3];
			cout<<endl;
		}
		cout<<endl;
	*/
	if (!is_ok()) return 0;
	if (x==1)
	{
		if (a[x2][y2]) return 1;
		else return 0;
	}
	for (int i=0;i<3;i++)
		for (int j=0;j<3;j++)
			if (a[i][j])
			{
				for (int l=0;l<8;l++)
					if (a[(i+c[l][0])%3][(j+c[l][1])%3]&&a[(i+2*c[l][0])%3][(j+2*c[l][1])%3]<b[(i+2*c[l][0])%3][(j+2*c[l][1])%3][l])
					{
						a[i][j]--;a[(i+c[l][0])%3][(j+c[l][1])%3]--;a[(i+2*c[l][0])%3][(j+2*c[l][1])%3]++;
						if (dfs(x-1)) return 1;
						a[i][j]++;a[(i+c[l][0])%3][(j+c[l][1])%3]++;a[(i+2*c[l][0])%3][(j+2*c[l][1])%3]--;

					}
			}
	return 0;
}
int main()
{
	freopen("galaxy.in","r",stdin);
	freopen("galaxy.out","w",stdout);
	while (scanf("%d%d%d%d%d",&k,&n,&m,&x2,&y2)!=EOF)
	{
		x2%=3;y2%=3;
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		for (int i=0;i<3;i++)
			for (int j=0;j<3;j++)
			{
				int h=((n/3)+(i>0)*(i<=n%3)),r=((m/3)+(j>0)*(j<=m%3));
				//b[i][j][8]=((n/3)+(i>0)*(i<=n%3))*((m/3)+(j>0)*(j<=m%3));
				b[i][j][8]=h*r;
				if (j!=0) b[i][j][0]=max(r-1,0)*h;else b[i][j][0]=r*h;
				if ((m+1)%3!=j) b[i][j][1]=max(r-1,0)*h;else b[i][j][1]=r*h;
				if (i!=0) b[i][j][2]=max(h-1,0)*r;else b[i][j][2]=r*h;
				if ((n+1)%3!=i) b[i][j][3]=max(h-1,0)*r;else b[i][j][3]=r*h;
				b[i][j][4]=max(r-(j!=0),0)*max(h-(i!=0),0);
				b[i][j][5]=max(r-((m+1)%3!=j),0)*max(h-((n+1)%3!=i),0);
				b[i][j][6]=max(r-((m+1)%3!=j),0)*max(h-(i!=0),0);
				b[i][j][7]=max(r-(j!=0),0)*max(h-((n+1)%3!=i),0);
			}

		for (int i=1;i<=k;i++)
		{
			int x,y;
			scanf("%d%d",&x,&y);
			a[x%3][y%3]++;
		}
		/*
		for (int j=3;j;j--)
		{
			for (int i=1;i<=3;i++)
				cout<<a[i%3][j%3];
			cout<<endl;
		}
		cout<<endl;
		*/
		if (dfs(k)) cout<<"Yesn";
		else cout<<"Non";
		h.clear();
	}
	return 0;
}