【算法】传教士和野人问题

有N个传教士和N个野人来到河边准备渡河，河岸有一条船，每次至多可供k人乘渡。问传教士为了安全起见，应如何规划摆渡方案，使得任何时刻，在河的两岸以及船上的野人数目总是不超过传教士的数目。即求解传教士和野人从左岸全部摆渡到右岸的过程中，任何时刻满足M（传教士数）≥C（野人数）和M＋C≤ k 的摆渡方案。

/*
 * Test.cpp
 *
 *  Created on: Aug 4, 2015
 *      Author: guangwei8.wu
 */

#include "stdafx.h"

#include 
#include 

#define SIZE  256

//condition nodes;
//ml, cl means scholars and cannibals on the left
//boat == 1 means left or right
//level to record times crossing the river
typedef struct
{
		int ml, cl, boat, level;
} conNode;

int N; //N the number of the cannibals  and also scholars
int M; //M	the limit of the boat
int Answer;
conNode conQueue[SIZE]; //N*N*2		condition queue
int flag[11][11][2]; //1 visited, 0 not

int front, rear;

void initQueue()
{
	front = 0;
	rear = 0;
}

int fullQueue()
{
	if (front == (rear + 1) % SIZE)
		return 1;
	else
		return 0;
}

int emptyQueue()
{
	if (front == rear)
		return 1;
	else
		return 0;
}
void inQueue(conNode node)
{
	if (fullQueue())
		return;
	conQueue[rear] = node;
	rear = (rear + 1) % SIZE;
}

conNode deQueue()
{
	if (!emptyQueue())
	{
		conNode n = conQueue[front];
		front = (front + 1) % SIZE;
		return n;
	}
}

int validCon(conNode n);

int main(void)
{
	int T, test_case;

	setbuf(stdout, NULL);
	scanf("%d", &T);
	for (test_case = 0; test_case < T; test_case++)
	{
		int f;
		conNode n;
		scanf("%d %d", &N, &M);

		Answer = -1;
		memset(flag, 0, sizeof(flag));
		initQueue();
		n.ml = N;
		n.cl = N;
		n.boat = 1;
		n.level = 0;
		inQueue(n);
		f = 1;
		flag[n.ml][n.cl][n.boat] = 1;
		while (!emptyQueue())
		{
			conNode tmp = deQueue();
			conNode newNode;
			int s = 0, c = 1;

			for (s = 0; s <= M; s++)
			{
				if (s == 0)
					c = 1;
				else
					c = 0;

				while ((s == 0 && s + c <= M) || (s != 0 && s + c <= M && c <= s))
				{
					if (tmp.boat == 1)
					{
						newNode.ml = tmp.ml - s;
						newNode.cl = tmp.cl - c;
						newNode.boat = 1 - tmp.boat;
						newNode.level = tmp.level + 1;
					}
					else
					{
						newNode.ml = tmp.ml + s;
						newNode.cl = tmp.cl + c;
						newNode.boat = 1 - tmp.boat;
						newNode.level = tmp.level + 1;
					}

					if (newNode.ml == 0 && newNode.cl == 0 && newNode.boat == 0)
					{
						f = 0;
						Answer = newNode.level;
						break;
					}

					if (validCon(newNode))
					{
						if (flag[newNode.ml][newNode.cl][newNode.boat] == 1)
						{
							c++;
							continue;
						}
						else
							flag[newNode.ml][newNode.cl][newNode.boat] = 1;

						inQueue(newNode);
					}

					c++;
				}

				if (f == 0)
					break;
			}
			if (f == 0)
				break;
		}
		if (Answer == -1)
			printf("impossible\n");
		else
			printf("%d\n", Answer);
	}

	return 0;
}

int validCon(conNode n)
{
	int ml = n.ml, cl = n.cl, boat = n.boat;

	if (ml != 0 && ml < cl)
		return 0;

	if (N - ml != 0 && N - ml < N - cl)
		return 0;

	if (ml < 0 || cl < 0)
		return 0;

	if (ml > N || cl > N)
		return 0;

	if (ml == N && cl == N && boat == 0)
		return 0;

	if (ml == 0 && cl == 0 && boat == 1)
		return 0;

	return 1;
}