BFS+Hash(储存，判重) HDOJ 1067 Gap

 

题目传送门

题意：一个图按照变成指定的图，问最少操作步数

分析；状态转移简单，主要是在图的存储以及判重问题，原来队列里装二维数组内存也可以，判重用神奇的hash技术

 

#include <bits/stdc++.h>
using namespace std;

const int MOD = 1e6 + 7;
struct Point	{
	int ch[5][9];
	int x[4], y[4];
	int step;
};
bool vis[MOD];
int ha;

int get_hash(int c[5][9])	{
	int tmp[60], k = 0;
	for (int i=1; i<=4; ++i)	{
		for (int j=2; j<=8; ++j)	{
			tmp[k++] = c[i][j] % 10;
			tmp[k++] = c[i][j] / 10;
		}
	}
	int ret = 0;
	for (int i=0; i<k; ++i)	{
		ret = (ret * 7 % MOD + tmp[i]) % MOD;
	}
	ret = (ret & 0x7fffffff) % MOD;
	return ret;
}

int init(void)	{
	int res[5][9];
	int x[4] = {11, 21, 31, 41};
	for (int i=0; i<4; ++i)	{
		for (int j=1; j<=7; ++j)	{
			res[i+1][j] = x[i]++;
		}
		res[i+1][8] = 0;
	}
	return get_hash (res);
}

void change(Point &v, int x0, int y0, int k)	{
	int a = v.ch[x0][y0-1] + 1;
	for (int i=1; i<=4; ++i)	{
		for (int j=2; j<=8; ++j)	{
			if (v.ch[i][j] == a)	{
				v.x[k] = i;	v.y[k] = j;
				swap (v.ch[x0][y0], v.ch[i][j]);
				return ;
			}
		}
	}
}

int BFS(Point &p)	{
	memset (vis, false, sizeof (vis));
	int sh = get_hash (p.ch);
	vis[sh] = true;
	queue<Point> que;	que.push (p);
	while (!que.empty ())	{
		Point u = que.front ();	que.pop ();
		int uh = get_hash (u.ch);
		if (uh == ha)	{
			return u.step;
		}
		for (int i=0; i<4; ++i)	{
			int x = u.x[i], y = u.y[i];
			if (y == 1 || u.ch[x][y] % 10 == 7)	continue;
			Point v = u;
			change (v, x, y, i);
			int vh = get_hash (v.ch);
			if (vis[vh])	continue;
			vis[vh] = true;	v.step++;
			que.push (v);
		}
	}

	return -1;
}

int main(void)	{
	ha = init ();	
	int T;	scanf ("%d", &T);
	while (T--)	{
		Point p;	p.step = 0;
		for (int i=1; i<=4; ++i)	{
			p.ch[i][1] = 0;
		}
		for (int i=1; i<=4; ++i)	{
			for (int j=2; j<=8; ++j)	{
				scanf ("%d", &p.ch[i][j]);
			}
		}
		for (int i=1; i<=4; ++i)	{
			for (int j=2; j<=8; ++j)	{
				if (p.ch[i][j] == 11)	{
					swap (p.ch[1][1], p.ch[i][j]);
					p.x[0] = i;	p.y[0] = j;
				}
				else if (p.ch[i][j] == 21)	{
					swap (p.ch[2][1], p.ch[i][j]);
					p.x[1] = i;	p.y[1] = j;
				}
				else if (p.ch[i][j] == 31)	{
					swap (p.ch[3][1], p.ch[i][j]);
					p.x[2] = i;	p.y[2] = j;
				}
				else if (p.ch[i][j] == 41)	{
					swap (p.ch[4][1], p.ch[i][j]);
					p.x[3] = i;	p.y[3] = j;
				}
			}
		}
		int ans = BFS (p);
		printf ("%d\n", ans);
	}

	return 0;
}