Pots POJ - 3414 （bfs+记录路径）

两杯水问题求解

最新推荐文章于 2020-03-06 16:23:53 发布

liuliu2333

最新推荐文章于 2020-03-06 16:23:53 发布

阅读量161

点赞数

CC 4.0 BY-SA版权

分类专栏： bfs

本文链接：https://blog.youkuaiyun.com/liuliu2333/article/details/81428564

bfs 专栏收录该内容

8 篇文章

订阅专栏

本文探讨了使用两种不同容量的容器如何通过一系列操作得到指定容量的水的问题。通过广度优先搜索算法寻找最短的操作路径，并详细展示了算法的具体实现过程。

You are given two pots, having the volume of A and B liters respectively. The following operations can be performed:

FILL(i) fill the pot i (1 ≤ i ≤ 2) from the tap;
DROP(i) empty the pot i to the drain;
POUR(i,j) pour from pot i to pot j; after this operation either the pot j is full (and there may be some water left in the pot i), or the pot i is empty (and all its contents have been moved to the pot j).

Write a program to find the shortest possible sequence of these operations that will yield exactly C liters of water in one of the pots.

Input

On the first and only line are the numbers A, B, and C. These are all integers in the range from 1 to 100 and C≤max(A,B).

Output

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
#include<stack>
#include<iostream>
using namespace std;
int a,b,c,vis[106][106];
struct nomd 
{
	int a,b,num,f;nomd *pre;
};
int ans=-1;
stack<int >r;
void bfs(){
	memset(vis,0,sizeof(vis));
	nomd e;nomd w[1000];
	e.a=0;e.b=0;e.num=0;e.f=0;e.pre=NULL;
	vis[0][0]=1;
	queue<nomd> que;
	que.push(e);
	int count=-1;
	while(!que.empty())
	{
		w[++count]=que.front();
		que.pop();
		if(w[count].a==c||w[count].b==c)
		{
			ans=w[count].num;
			while(w[count].pre)
			{
				r.push(w[count].f);
				w[count]= *w[count].pre;
			}
			break;
		}
		for(int i=1;i<=6;i++)
		{
		   e=w[count];
			if(i==1)
			{
			 e.a=a;
			}
			if(i==2){
				e.b=b;
			}
			if(i==3)
			{
			  e.a=0;
			}
			if(i==4){
				e.b=0;
			}
			if(i==5){
				e.a=w[count].a-(b-w[count].b);e.b=w[count].a+w[count].b;
				if(e.b>b)  e.b=b;
				if(e.a<0) e.a=0;
			}
			if(i==6){
				e.b=w[count].b-(a-w[count].a);e.a=w[count].a+w[count].b;
				if(e.a>a) e.a=a;
				if(e.b<0) e.b=0;
			}
			if(vis[e.a][e.b]==1) continue;
			e.f=i; e.pre=&w[count];
			e.num=w[count].num+1;
			vis[e.a][e.b]=1;
			que.push(e);
		}
		
	}
}
int main()
{

	scanf("%d %d %d",&a,&b,&c);
	bfs();
//	for(int i=0;i<=a;i++)
//	{
//		for(int j=0;j<=b;j++)
//		printf("%d ",vis[i][j]);
//		printf("\n");
//	}
	if(ans==-1)
	{
		printf("impossible\n");
	}
	else
	printf("%d\n",ans);
	while(!r.empty())
	{
		int i = r.top();
		r.pop();
		switch(i)
		{
			case 1:cout<<"FILL(1)"<<endl;break;
			case 2:cout<<"FILL(2)"<<endl;break;
			case 3:cout<<"DROP(1)"<<endl;break;
			case 4:cout<<"DROP(2)"<<endl;break;
			case 5:cout<<"POUR(1,2)"<<endl;break;
			case 6:cout<<"POUR(2,1)"<<endl;break;
		}
	}
	return 0;
}

The first line of the output must contain the length of the sequence of operations K. The following K lines must each describe one operation. If there are several sequences of minimal length, output any one of them. If the desired result can’t be achieved, the first and only line of the file must contain the word ‘impossible’.

Sample Input

3 5 4

Sample Output

6
FILL(2)
POUR(2,1)
DROP(1)
POUR(2,1)
FILL(2)
POUR(2,1)

这道题的意思是两个杯子6种操作，1.将a装满 2.将b装满 3.将a倒光 4.将b倒光 5.将a倒入b 6 将b倒入a.

主要是记录路径