BFS(八数码) POJ 1077 || HDOJ 1043 Eight

 

题目传送门1 2

题意:从无序到有序移动的方案,即最后成1 2 3 4 5 6 7 8 0

分析:八数码经典问题。POJ是一次,HDOJ是多次。因为康托展开还不会,也写不了什么,HDOJ需要从最后的状态逆向搜索,这样才不会超时。判重康托展开,哈希也可。

 

POJ

//#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<stack>
#include<queue>
#include <cstring>
#include<map>
#include<stdio.h>
#include<stdlib.h>
#include<ctype.h>
#include<time.h>
#include<math.h>
using namespace std;

const int N = 362880 + 5;
const int MOD = 1e6 + 7;
int dx[4] = {-1, 1, 0, 0};
int dy[4] = {0, 0, -1, 1};
char dir[4] = {'u', 'd', 'l', 'r'};
struct Point	{
	int s, d;
	string str;
	Point () {}
	Point (int s, int d, string str) : s (s), d (d), str (str) {}
};
struct Hash_table	{
	struct Edge	{
		int v, nex;
	}edge[MOD];
	int head[MOD], e;
	void init(void)	{
		memset (head, -1, sizeof (head));
		e = 0;
	}
	bool insert(int x)	{
		int u = (x % MOD + MOD) % MOD;
		for (int i=head[u]; ~i; i=edge[i].nex)	{
			if (edge[i].v == x)	return false;
		}
		edge[e].v = x;	edge[e].nex = head[u];
		head[u] = e++;
		return true;
	}
}ha;
int vis[N], fact[9];

void decode(int x, int *b)	{
	for (int i=8; i>=0; --i)	{
		b[i] = x % 10;
		x /= 10;
	}
}

int encode(int *b)	{
	int ret = 0;
	for (int i=0; i<9; ++i)	{
		ret = ret * 10 + b[i];
	}
	return ret;
}

int find_0(int *b)	{
	for (int i=0; i<9; ++i)	{
		if (b[i] == 0)	return i;
	}
	return -1;
}

bool check(int x, int y)	{
	if (x < 0 || x >= 3 || y < 0 || y >= 3)	return false;
	else	return true;
}

void print(int *b)	{
	for (int i=0; i<9; ++i)	{
		printf ("%d ", b[i]);
		if (i == 2 || i == 5 || i == 8)	puts ("");
	}
}

void init(void)	{
	fact[0] = 1;
	for (int i=1; i<9; ++i)	fact[i] = fact[i-1] * i;
	memset (vis, false, sizeof (vis));
}

bool can_insert(int *b)	{
	int code = 0;
	for (int i=0; i<9; ++i)	{
		int cnt = 0;
		for (int j=i+1; j<9; ++j)	if (b[j] < b[i])	cnt++;
		code += fact[8-i] * cnt;
	}
	if (vis[code])	return false;
	else	{
		vis[code] = true;
		return true;
	}
}

void BFS(int *a)	{
	init ();
	int ans[9] = {1, 2, 3, 4, 5, 6, 7, 8, 0};
	int s = encode (a);
	queue<Point> que;	que.push (Point (s, 0, ""));
	while (!que.empty ())	{
		Point u = que.front ();	que.pop ();
		int b[9];
		decode (u.s, b);
		if (memcmp (ans, b, sizeof (b)) == 0)	{
			int len = u.str.length ();
			for (int i=0; i<len; ++i)	{
				printf ("%c", u.str[i]);
			}
			puts ("");
			return ;
		}
		int p = find_0 (b);
		int x = p / 3, y = p % 3;
		for (int i=0; i<4; ++i)	{
			int tx = x + dx[i], ty = y + dy[i];
			if (!check (tx, ty))	continue;
			int p2 = tx * 3 + ty;
			int t[9];
			memcpy (t, b, sizeof (b));
			t[p] = t[p2];	t[p2] = 0;
			int v = encode (t);
			//if (!ha.insert (v))	continue;
			if (!can_insert (t))	continue;
			que.push (Point (v, u.d + 1, u.str + dir[i]));
		}
	}
	puts ("unsolvable");
}

int main(void)	{
	char c[9];
	int a[9];
	while (scanf ("%c %c %c %c %c %c %c %c %c", &c[0], &c[1], &c[2], &c[3], &c[4], &c[5], &c[6], &c[7], &c[8]) == 9)	{
		for (int i=0; i<9; ++i)	{
			if (c[i] >= '1' && c[i] <= '9')	{
				a[i] = c[i] - '0';
			}
			else	a[i] = 0;
		}
		BFS (a);
	}

	return 0;
}

  

HDOJ

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

const int N = 362880 + 5;
int dx[4] = {-1, 1, 0, 0};
int dy[4] = {0, 0, -1, 1};
char dir[4] = {'d', 'u', 'r', 'l'};
struct Point	{
	int s;
	int b[9];
};
struct Ans	{
	char dir;
	int fa;
}ans[N];
int fact[9];

int find_0(int *b)	{
	for (int i=0; i<9; ++i)	{
		if (b[i] == 9)	return i;
	}
	return -1;
}

bool check(int x, int y)	{
	if (x < 0 || x >= 3 || y < 0 || y >= 3)	return false;
	else	return true;
}

void init(void)	{
	fact[0] = 1;
	for (int i=1; i<9; ++i)	fact[i] = fact[i-1] * i;
	for (int i=0; i<N; ++i)	ans[i].fa = -1;
}

int Cantor(int *b)	{
	int code = 0;
	for (int i=0; i<9; ++i)	{
		int cnt = 0;
		for (int j=i+1; j<9; ++j)	if (b[j] < b[i])	cnt++;
		code += fact[8-i] * cnt;
	}
	return code;
}

void BFS()	{
	init ();
	Point sta;
	for (int i=0; i<9; ++i)	{
		sta.b[i] = i + 1;
	}
	sta.s = 0;	ans[sta.s].fa = 0;
	queue<Point> que;	que.push (sta);
	while (!que.empty ())	{
		Point u = que.front ();	que.pop ();
		int p = find_0 (u.b);
		int x = p / 3, y = p % 3;
		for (int i=0; i<4; ++i)	{
			Point v = u;
			int tx = x + dx[i], ty = y + dy[i];
			if (!check (tx, ty))	continue;
			int p2 = tx * 3 + ty;
			swap (v.b[p], v.b[p2]);
			v.s = Cantor (v.b);
			if (ans[v.s].fa != -1)	continue;
			ans[v.s].dir = dir[i];
			ans[v.s].fa = u.s;
			que.push (v);
		}
	}
}

int main(void)	{
	BFS ();
	char c[55];
	int a[9];
	while (gets (c))	{
		int j = 0;
		for (int i=0; c[i]; ++i)	{
			if (c[i] >= '0' && c[i] <= '8')	{
				a[j++] = c[i] - '0';
			}
			else if (c[i] == 'x')	a[j++] = 9;
		}
		int s = Cantor (a);
		if (ans[s].fa == -1)	puts ("unsolvable");
		else	{
			while (s != 0)	{
				printf ("%c", ans[s].dir);
				s = ans[s].fa;
			}
			puts ("");
		}
	}

	return 0;
}

  

    原文作者:Running_Time
    原文地址: https://www.cnblogs.com/Running-Time/p/4984695.html
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
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