空瓶子装水解谜游戏

<pre name="code" class="cpp">#include <iostream>
#include <queue>
#include <math.h>
using namespace std;

#define N 1000000
#define M 10
int array[N][M];
int bfs(int *a, int cur, int x, int d , int n)
{
	if (cur == d)
		return 0;
	for (int i = 0; i < n; i++)
	{
		if (cur == 0 || array[cur-1][i] < a[i])
		{
			if (cur)
				for (int k = 0; k < n; k++)
				{
					array[cur][k] = array[cur - 1][k];
				}
			else 
				for (int k = 0; k < n; k++)
				{
					array[cur][k] = 0;
				}
			array[cur][i] = a[i];
			if (array[cur][0] == x)
			{
				for (int m = 0; m <= cur; m++)
				{
					for (int k = 0; k <= n; k++)
						cout << array[m][k] << ' ';
					cout << endl;
				}
				return 1;
			}
			else
			{
				if (bfs(a, cur + 1, x, d, n)) return 1;
			}
		}
		if (cur && array[cur-1][i] > 0)
		{
			for (int k = 0; k < n; k++)
			{
				array[cur][k] = array[cur - 1][k];
			}
			array[cur][i] = 0;
			if (array[cur][0] == x)
			{
				for (int m = 0; m <= cur; m++)
				{
					for (int k = 0; k < n; k++)
						cout << array[m][k] << ' ';
					cout << endl;
				}
				return 1;
			}
			else
			{
				if (bfs(a, cur + 1, x, d, n))	return 1;	
			}
		}
		if (cur)
		{
			for (int j = 0; j < n; j++)
			{
				for (int k = 0; k < n; k++)
				{
					array[cur][k] = array[cur - 1][k];
				}
				if (i != j)
				{
					if (array[cur - 1][j] < a[j] && array[cur - 1][i] > 0)//此时可以倒水i给j倒水
					{
						array[cur][i] = array[cur - 1][i] + array[cur - 1][j] <= a[j] ? 0 : array[cur - 1][i] + array[cur - 1][j] - a[j];
						array[cur][j] = array[cur - 1][i] + array[cur - 1][j] <= a[j] ? array[cur - 1][i] + array[cur - 1][j] : a[j];
						if (array[cur][0] == x)// || array[cur][j] == x)
						{
							for (int m = 0; m <= cur; m++)
							{
								for (int k = 0; k < n; k++)
									cout << array[m][k] << ' ';
								cout << endl;
							}
							return 1;
						}
						else
						{
							if (bfs(a, cur + 1, x, d, n)) return 1;
						}
					}
				}
			}
		}
	}
	return 0;
}

int main()
{
	int a[M], n,x;
	while (1)
	{
		cin >> n;
		for (int i = 0; i < n; i++)
			cin >> a[i];
		cin >> x;
		for (int i = 1;; i++)
		{
			for (int j = 0; j < n; j++)
				array[0][j] = 0;
			if (bfs(a, 0, x, i, n))
				break;
		}
	}
	return 0;
}




                
/* * 基于双向链表实现双端队列结构 */ package dsa; public class Deque_DLNode implements Deque { protected DLNode header;//指向头节点(哨兵) protected DLNode trailer;//指向尾节点(哨兵) protected int size;//队列中元素的数目 //构造函数 public Deque_DLNode() { header = new DLNode(); trailer = new DLNode(); header.setNext(trailer); trailer.setPrev(header); size = 0; } //返回队列中元素数目 public int getSize() { return size; } //判断队列是否为 public boolean isEmpty() { return (0 == size) ? true : false; } //取首元素(但不删除) public Object first() throws ExceptionQueueEmpty { if (isEmpty()) throw new ExceptionQueueEmpty("意外:双端队列为"); return header.getNext().getElem(); } //取末元素(但不删除) public Object last() throws ExceptionQueueEmpty { if (isEmpty()) throw new ExceptionQueueEmpty("意外:双端队列为"); return trailer.getPrev().getElem(); } //在队列前端插入新节点 public void insertFirst(Object obj) { DLNode second = header.getNext(); DLNode first = new DLNode(obj, header, second); second.setPrev(first); header.setNext(first); size++; } //在队列后端插入新节点 public void insertLast(Object obj) { DLNode second = trailer.getPrev(); DLNode first = new DLNode(obj, second, trailer); second.setNext(first); trailer.setPrev(first); size++; } //删除首节点 public Object removeFirst() throws ExceptionQueueEmpty { if (isEmpty()) throw new ExceptionQueueEmpty("意外:双端队列为"); DLNode first = header.getNext(); DLNode second = first.getNext(); Object obj = first.getElem(); header.setNext(second); second.setPrev(header); size--; return(obj); } //删除末节点 public Object removeLast() throws ExceptionQueueEmpty { if (isEmpty()) throw new ExceptionQueueEmpty("意外:双端队列为"); DLNode first = trailer.getPrev(); DLNode second = first.getPrev(); Object obj = first.getElem(); trailer.setPrev(second); second.setNext(trailer); size--; return(obj); } //遍历 public void Traversal() { DLNode p = header.getNext(); while (p != trailer) { System.out.print(p.getElem()+" "); p = p.getNex
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