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2013
12-10

HDU 1415 Jugs-BFS-[解题报告] C++

Jugs

问题描述 :

In the movie "Die Hard 3", Bruce Willis and Samuel L. Jackson were confronted with the following puzzle. They were given a 3-gallon jug and a 5-gallon jug and were asked to fill the 5-gallon jug with exactly 4 gallons. This problem generalizes that puzzle.

You have two jugs, A and B, and an infinite supply of water. There are three types of actions that you can use: (1) you can fill a jug, (2) you can empty a jug, and (3) you can pour from one jug to the other. Pouring from one jug to the other stops when the first jug is empty or the second jug is full, whichever comes first. For example, if A has 5 gallons and B has 6 gallons and a capacity of 8, then pouring from A to B leaves B full and 3 gallons in A.

A problem is given by a triple (Ca,Cb,N), where Ca and Cb are the capacities of the jugs A and B, respectively, and N is the goal. A solution is a sequence of steps that leaves exactly N gallons in jug B. The possible steps are

fill A
fill B
empty A
empty B
pour A B
pour B A
success

where "pour A B" means "pour the contents of jug A into jug B", and "success" means that the goal has been accomplished.

You may assume that the input you are given does have a solution.

输入:

Input to your program consists of a series of input lines each defining one puzzle. Input for each puzzle is a single line of three positive integers: Ca, Cb, and N. Ca and Cb are the capacities of jugs A and B, and N is the goal. You can assume 0 < Ca <= Cb and N <= Cb <=1000 and that A and B are relatively prime to one another.

输出:

Output from your program will consist of a series of instructions from the list of the potential output lines which will result in either of the jugs containing exactly N gallons of water. The last line of output for each puzzle should be the line "success". Output lines start in column 1 and there should be no empty lines nor any trailing spaces.

样例输入:

3 5 4 
5 7 3 

样例输出:

fill B 
pour B A 
empty A 
pour B A 
fill B 
pour B A 
success 
fill A 
pour A B 
fill A 
pour A B 
empty B 
pour A B 
success 

题目链接:

题目大意:这题是倒水问题,现在有两个容积为a和b的水壶,对每个水壶可以进行4种操作,两个水壶之间相互倒水(一个水壶倒空或者一个水壶倒满为止),从水农头那里灌水(将水壶灌满为止),向外倒水(将水壶倒空为止),问对这两个水壶进行这样的一系列操作是否可以量出容积为c的水(两个杯子中有一个水壶中的水的容积恰好为c)
     这里添加一个水壶编号为0,容积为a和b的水壶分别编号为1和2,编号为0的水壶的容积置为a+b(保证0号水壶可以向1或2中加水,或者1或2向0中加水),再用BFS来进行求解.

代码如下:

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=110;
int sh[maxn],result;
int vis[maxn][maxn];
struct node
{
	int v[3];
	int fa;
}q[maxn*maxn];
node path[maxn*maxn];
void Print_path(int ans)
{
	int i=0,j,k;
	node p1,p2;
	while(q[ans].fa!=ans)
	{
		path[i++]=q[ans];
		ans=q[ans].fa;
	}
	path[i]=q[ans];
	while(i>=1)
	{
		p1=path[i];
		p2=path[--i];
		if(p1.v[0]==p2.v[0])//用标号为0的水壶中德前后两个状态作为判断标准
		{
			if(p1.v[1]<p2.v[1])
				printf("pour B A\n");
			else
				printf("pour A B\n");
		}
		else
		{
			for(j=1;j<3;j++)
				if(p1.v[j]==p2.v[j])
					break;
			k=3-j;
			if(p1.v[k]<p2.v[k])
				printf("fill %c\n",k-1+'A');
			else
				printf("empty %c\n",k-1+'A');
		}
	}
	printf("success\n\n");
}
void bfs()
{
	int i,j,k,start=0,tail=1;
	int amount;
	q[0].v[0]=sh[0];
	q[0].v[1]=q[0].v[2]=q[0].fa=0;
	vis[0][0]=1;
	while(start<tail)
	{
		node &u=q[start];
		if(u.v[1]==result||u.v[2]==result)
		{
			Print_path(start);//到达目标,打印方案
			return;
		}
		for(i=0;i<3;i++)//标号为i的水壶向标号为j的水壶中倒水
			for(j=0;j<3;j++)
				if(i!=j)
				{
					node &v=q[tail];
					if(u.v[i]<sh[j]-u.v[j])
						amount=u.v[i];//amount是此次操作可以倒出的水量
					else
						amount=sh[j]-u.v[j];
					for(k=0;k<3;k++)
						v.v[k]=u.v[k];//扩展新节点
					v.v[i]-=amount;//倒出
					v.v[j]+=amount;//倒进
					if(!vis[v.v[1]][v.v[2]])
					{
						vis[v.v[1]][v.v[2]]=1;//标记为已访问
						v.fa=start;
						tail++;
					}
				}
				start++;
	}
}
int main(void)
{
	while(scanf("%d%d%d",&sh[1],&sh[2],&result)!=EOF)
	{
		sh[0]=sh[1]+sh[2];
		memset(vis,0,sizeof(vis));
		bfs();
	}
}

 

解题报告转自:http://blog.csdn.net/chao_xun/article/details/8040071


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