【BFS|位运算+输出路径】POJ-2965 The Pilots Brothers' refrigerator

最新推荐文章于 2019-12-03 14:50:36 发布

J_Sure

最新推荐文章于 2019-12-03 14:50:36 发布

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分类专栏：位运算 ACM-POJ BFS 文章标签： acm c++ oj

本文链接：https://blog.youkuaiyun.com/j_sure/article/details/41039031

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本文介绍了一款游戏中关于打开特殊冰箱门的谜题解决方法。该冰箱有16个开关，通过位压缩BFS算法找到最少操作次数及具体步骤来全部开启。文章详细解释了解题思路，并附上了实现代码。

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The Pilots Brothers' refrigerator

Time Limit: 1000MS		Memory Limit: 65536K
				Special Judge

Description

The game “The Pilots Brothers: following the stripy elephant” has a quest where a player needs to open a refrigerator.

There are 16 handles on the refrigerator door. Every handle can be in one of two states: open or closed. The refrigerator is open only when all handles are open. The handles are represented as a matrix 4х4. You can change the state of a handle in any location [i, j] (1 ≤ i, j ≤ 4). However, this also changes states of all handles in row iand all handles in column j.

The task is to determine the minimum number of handle switching necessary to open the refrigerator.

Input

The input contains four lines. Each of the four lines contains four characters describing the initial state of appropriate handles. A symbol “+” means that the handle is in closed state, whereas the symbol “−” means “open”. At least one of the handles is initially closed.

Output

The first line of the input contains N – the minimum number of switching. The rest N lines describe switching sequence. Each of the lines contains a row number and a column number of the matrix separated by one or more spaces. If there are several solutions, you may give any one of them.

Sample Input

-+--
----
----
-+--

Sample Output

————————————————————悲しみの分割線————————————————————

前言：又被学弟抢先A了，血虐。。。悲哀。。。我天天在干嘛？！？？！？？！一个星期才做了几题？？！？？！？！？打脸！啪！。。。

思路：只是一道位压缩的BFS而已，做了这么久。路径输出本来就不难，我竟然SB地记录枚举的路径！那不肯定是0～15吗？！？？？！把队列转移路径记录下来！

代码如下：

/*
ID: j.sure.1
PROG:
LANG: C++
*/
/****************************************/
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <ctime>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <climits>
#include <iostream>
#define LL long long
using namespace std;
const int INF = 0x3f3f3f3f;
/****************************************/
const int N = 1e5;
struct Node {
	int m, step, dir;
	Node(){}
	Node(int _m, int _s, int _d):
		m(_m), step(_s), dir(_d){}
};
Node q[N];
bool vis[0xFFFF+1];
int flag, fa[N];

Node pull(Node t, int i)
{
	Node u;
	int x = i >> 2, y = i & 3;//取出行和列
	//0123 4567 89AB CDEF
	x <<= 2;
	for(int k = 0; k < 4; k++) {
		t.m ^= 1 << (x+k);
	}
	for(int k = 0; k <= 12; k+=4) {
		t.m ^= 1 << (y+k);
	}
	t.m ^= 1 << i;
	u.m = t.m;
	u.dir = i;
	u.step = t.step + 1;
	return u;
}

int bfs()
{
	int fron = 0, rear = 1; 
	while(fron < rear) {
		Node t = q[fron];
		if(!t.m) {
			flag = fron;
			return t.step;
		}
		for(int i = t.dir + 1; i < 16; i++) {
			Node u = pull(t, i);
			if(!vis[u.m]) {
				vis[u.m] = 1;
				fa[rear] = fron;
				q[rear++] = u;
			}
		}
		fron++;
	}
	return -1;
}

int main()
{
#ifdef J_Sure
//	freopen("000.in", "r", stdin);
//	freopen(".out", "w", stdout);
#endif
	char s[4][10];
	int x = 0;
	for(int i = 0; i < 4; i++) {
		scanf("%s", s[i]);
	}
	for(int i = 3; i >= 0; i--) {
		for(int j = 3; j >= 0; j--) {
			if(s[i][j] == '+') {
				x = x << 1 | 1;
			}
			else x <<= 1;
		}
	}
	vis[x] = 1;
	q[0] = Node(x, 0, -1);
	fa[0] = -1;
	int ans = bfs();
	printf("%d\n", ans);
	while(~fa[flag]) {
		int tmp = q[flag].dir;
		printf("%d %d\n", (tmp>>2)+1, (tmp&3)+1);
		flag = fa[flag];
	}
	return 0;
}