poj 3414 pots（bfs暴力+路径记录）

最新推荐文章于 2020-09-10 23:35:31 发布

摆渡过江

最新推荐文章于 2020-09-10 23:35:31 发布

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CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/qq_39562952/article/details/82829228

博客文章介绍了如何解决POJ 3414 Pots问题，主要采用BFS广度优先搜索，并通过记录路径来解决六种情况的排列组合。作者通过构建图结构存储已访问状态，每个节点包含父节点位置及到达方式。若找到逆序排列则输出，否则输出"impossible"。文章提供详细代码实现。

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思路：这可真是我目前写过最长的题目了，虽然很多可以复制（学艺不精T-T）。。。

其实思路还是很简单的，就是6种情况，各种倒来倒去，然后记录下路径，我的方法是用一张图来表示，有没有走到过，然后图上每个位置都存着父节点的位置，以及父节点怎么达到，如果找到了逆序输出，如果没找到impossible，具体的看代码吧；

代码如下：

#include <iostream>
#include <string>
#include <vector>
#include <set>
#include <cstring>
#include <climits>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stdio.h>
#include <sstream>
#define esp 1e-4
using namespace std;
int A, B, C;
int ans;
struct node
{
	int x, y;//代表现在1号杯子跟2号杯子的状态
	int fx, fy;//代表上一个状态
	string to;//代表从上一个状态怎么到当前的状态 
	int step;//到达的时间 
};
node chart[110][110];
bool vis[110][110];
int judge(int x, int y)
{
	if (x == C || y == C)
		return 1;
	else
		return 0;
}
void solve(int x, int y, int step)
{
	ans = step;
	cout << step << endl;
	string temp[10000];
	int time = 0;
	while (step--)
	{
		node kkk=chart[x][y];
		temp[time] = kkk.to;
		x = kkk.fx;
		y = kkk.fy;
		time++;
	}
	for (int i =ans - 1; i >= 0; i--)
	{
		cout << temp[i] << endl;
	}
	
}
void bfs()
{
	node temp;
	queue<node> q;
	temp.x = 0;
	temp.y = 0;
	temp.step = 0;
	vis[0][0] = true;
	q.push(temp);
	while (!q.empty())
	{
		node then;
		//倒空1号瓶
		if (!vis[0][q.front().y])
		{
			then.fx = q.front().x;
			then.fy = q.front().y;
			then.to = "DROP(1)";
			then.x = 0;
			then.y = q.front().y;
			then.step = q.front().step + 1;
			vis[then.x][then.y] = true;
			chart[then.x][then.y].fx = q.front().x;
			chart[then.x][then.y].fy = q.front().y;
			chart[then.x][then.y].to = "DROP(1)";
			if (judge(then.x, then.y))
			{
				solve(then.x, then.y, then.step);
				break;
			}
			
			q.push(then);
		}
		//倒空2号瓶
		if (!vis[q.front().x][0])
		{
			then.fx = q.front().x;
			then.fy = q.front().y;
			then.to = "DROP(2)";
			then.y = 0;
			then.x = q.front().x;
			then.step = q.front().step + 1;
			vis[then.x][then.y] = true;
			chart[then.x][then.y].fx = q.front().x;
			chart[then.x][then.y].fy = q.front().y;
			chart[then.x][then.y].to = "DROP(2)";
			if (judge(then.x, then.y))
			{
				solve(then.x, then.y, then.step);
				break;
			}
			
			q.push(then);
		}
		//倒满1号瓶
		if (!vis[A][q.front().y])
		{
			then.fx = q.front().x;
			then.fy = q.front().y;
			then.to = "FILL(1)";
			then.x = A;
			then.y = q.front().y;
			then.step = q.front().step + 1;
			vis[then.x][then.y] = true;
			chart[then.x][then.y].fx = q.front().x;
			chart[then.x][then.y].fy = q.front().y;
			chart[then.x][then.y].to = "FILL(1)";
			if (judge(then.x, then.y))
			{
				solve(then.x, then.y, then.step);
				break;
			}
			
			q.push(then);
		}
		//倒满2号瓶
		if (!vis[q.front().x][B])
		{
			then.fx = q.front().x;
			then.fy = q.front().y;
			then.to = "FILL(2)";
			then.x = q.front().x;
			then.y = B;
			then.step = q.front().step + 1;
			vis[then.x][then.y] = true;
			chart[then.x][then.y].fx = q.front().x;
			chart[then.x][then.y].fy = q.front().y;
			chart[then.x][then.y].to = "FILL(2)";
			if (judge(then.x, then.y))
			{
				solve(then.x, then.y, then.step);
				break;
			}
			
			q.push(then);
		}
		//把1号瓶倒入2号瓶 
		if (q.front().x + q.front().y <= B)//如果倒的下 
		{
			if (!vis[0][q.front().x + q.front().y])
			{
				then.fx = q.front().x;
				then.fy = q.front().y;
				then.step = q.front().step + 1;
				then.to = "POUR(1,2)";
				then.x = 0;
				then.y = q.front().x + q.front().y;
				vis[then.x][then.y] = true;
				chart[then.x][then.y].fx = q.front().x;
				chart[then.x][then.y].fy = q.front().y;
				chart[then.x][then.y].to = "POUR(1,2)";
				if (judge(then.x, then.y))
				{
					solve(then.x, then.y, then.step);
					break;
				}
				
				q.push(then);
			}
			
		}
		else//如果倒不下 
		{
			if (!vis[q.front().x + q.front().y - B][B])
			{
				then.fx = q.front().x;
				then.fy = q.front().y;
				then.step = q.front().step + 1;
				then.to = "POUR(1,2)";
				then.x = q.front().x + q.front().y - B;
				then.y = B;
				vis[then.x][then.y] = true;
				chart[then.x][then.y].fx = q.front().x;
				chart[then.x][then.y].fy = q.front().y;
				chart[then.x][then.y].to = "POUR(1,2)";
				if (judge(then.x, then.y))
				{
					solve(then.x, then.y, then.step);
					break;
				}
				
				q.push(then);
			}
			
		}
		//把2号瓶倒入1号瓶
		if (q.front().x + q.front().y <= A)//到的下 
		{
			if (!vis[q.front().x + q.front().y][0])
			{
				then.fx = q.front().x;
				then.fy = q.front().y;
				then.step = q.front().step + 1;
				then.to = "POUR(2,1)";
				then.x = q.front().x + q.front().y;
				then.y = 0;
				vis[then.x][then.y] = true;
				chart[then.x][then.y].fx = q.front().x;
				chart[then.x][then.y].fy = q.front().y;
				chart[then.x][then.y].to = "POUR(2,1)";
				if (judge(then.x, then.y))
				{
					solve(then.x, then.y, then.step);
					break;
				}
				
				q.push(then);
			}
			
		}
		else
		{
			if (!vis[A][q.front().x + q.front().y - A])
			{
				then.fx = q.front().x;
				then.fy = q.front().y;
				then.step = q.front().step + 1;
				then.to = "POUR(2,1)";
				then.x = A;
				then.y = q.front().x + q.front().y - A;
				vis[then.x][then.y] = true;
				chart[then.x][then.y].fx = q.front().x;
				chart[then.x][then.y].fy = q.front().y;
				chart[then.x][then.y].to = "POUR(2,1)";
				if (judge(then.x, then.y))
				{
					solve(then.x, then.y, then.step);
					break;
				}
				
				q.push(then);
			}
			
		}
		//cout << q.front().fx << q.front().fy << q.front().step << q.front().to << q.front().x << q.front().y << endl;
		q.pop();

	}
}
int main()
{
	while (cin >> A >> B >> C)
	{
		memset(vis, 0, sizeof(vis));
		ans = -1;
		bfs();
		if (ans == -1)
		{
			cout << "impossible\n";
		}
	}
	/*
	3 5 4

	*/
	return 0;
}