2010-09-21北大创新工场笔试大题_北大上海临港科创中心笔试题-优快云博客

本文介绍了一个经典的两容器量水问题，并提供了一种基于回溯法的解决方案。该方案通过递归尝试不同的倒水操作来找出如何从两个不同容量的容器中准确得到目标水量的方法。

题目：

有两个量杯A和B，分别可以量M和N升水，还有装满水的池子，以及一个大的空杯子C，现在要经过一系列的步骤在空杯子里面量出K升水。编写程序，输入为M，N，K，输出为步骤，或者“No Solution”。例如M, N, K分别为5, 7, 4，则输出的步骤为：

0, 0, 0 (都没有水)

0, 7, 0 (B倒满水)

5, 2, 0 (B倒入A)

5, 0, 2 (B到入C)

0, 0, 2 (B倒入水池)

0, 7, 2 (B倒满水)

5, 2, 2 (B倒入A)

5, 0, 4 (B到入C，达到要求)

采用回溯法

#include <iostream>
#include <stack>
#include <cstdlib>
using namespace std;
enum Action
{
	ACTION_A,
	ACTION_B,
	ACTION_AB,
	ACTION_A2B,
};
static void print_action(const int M, const int N, int ∑,
		const Action a, const bool a_larger)
{
	switch (a){
	case ACTION_A:
		cout << M << ", 0, " << sum << endl; 
		sum += M;
		cout << 0 << ", 0, " << sum << endl; 
		break;
	case ACTION_B:
		cout << "0, " << N << ", " << sum << endl; 
		sum += N;
		cout << "0, 0, " << sum << endl; 
		break;
	case ACTION_AB:
		if (a_larger){
			cout << M << ", 0, " << sum << endl; 
			cout << M - N << ", " << N << ", " << sum << endl; 
			sum += (M - N);
			cout << "0, " << N << ", " << sum << endl; 
		}else{
			cout << "0, " << N << ", " << sum << endl; 
			cout << M << ", " << N - M << ", " << sum << endl; 
			sum += (N - M);
			cout << M << ", " << 0 << ", " << sum << endl; 
		}
		cout << "0, 0, " << sum << endl;
		break;
	case ACTION_A2B:
		break;
	default:
		cout << "error in print_action" << endl;
		exit(-1);
	}
}
static void print_solution(const int M, const int N,
		stack<Action> &action, const bool a_larger)
{
	Action a;
	int sum = 0;
	while (!action.empty()){
		a = action.top();
		print_action(M, N, sum, a, a_larger);
		action.pop();
	}
}
static void try_to_fill(const int M, const int N, const int K, 
		stack<Action> &action, const bool a_larger)
{
	if (0 == K){
		print_solution(M, N, action, a_larger);
		exit(0);
	}else if (K < 0){
		return;
	}
	action.push(ACTION_A);
	try_to_fill(M, N, K - M, action, a_larger);
	action.pop();
	action.push(ACTION_B);
	try_to_fill(M, N, K - N, action, a_larger);
	action.pop();
	action.push(ACTION_AB);
	try_to_fill(M, N, K - (a_larger ? (M - N) : (N - M)), action, a_larger);
	action.pop();
}

static void simple_fill(const int M, const int K, stack<Action> &action)
{
	int tmp = K;
	while (tmp > 0){
		action.push(ACTION_A);
		tmp -= M;
	}
	if (0 == tmp){
		print_solution(M, M, action, 0);
	}else{
		cout << "No Solution" << endl;
	}
}
int main()
{
	const int M = 5, N = 7, K = 4;
	stack<Action> action;
	bool a_larger = (M > N);
	cout << M << ", " << N << ", " << K << endl;
	cout << "******************" << endl;
	cout << "0, 0, 0" << endl; /*all empty*/
	if (M == N){
		simple_fill(M, K, action);
		return 0;
	}
	try_to_fill(M, N, K, action, a_larger);
	action.push(ACTION_A2B);
	try_to_fill(M, N, K - (a_larger ? (2*N - M) : (2*M - N)), 
			action, a_larger);
	cout << "No Solution" << endl;
	return 0;
}