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SDUT--Pots--BFS

2024-07-17 04:17:05   228人阅读

Pots

Time Limit: 1000ms   Memory limit: 65536K  有疑问？点这里^_^

题目描述

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.

输入

 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).

输出

 The first line of the output must contain the length of the sequence of operations K.  If the desired result can’t be achieved, the first and only line of the file must contain the word ‘impossible’.

示例输入

3 5 4

示例输出

6

提示

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

简单搜索题，对于两个空瓶子，容积分别为A B 有6种操作 把A（或B）清空，把A（或B）装满，把A倒入B，把B倒入A 。对应这6种操作，有6种状态，典型的bfs搜索。不多了，只是这题明明说的是单组输入结果答案却要多组输入才对，白白贡献5个WA。
#include <cstdio>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <queue>
using namespace std;
int m,n,c;
typedef struct node
{
	int v1,v2,op;
};
bool vis[999][999];
void bfs()
{
	node t={0,0,0};
	queue <node> Q;
	Q.push(t);
	vis[0][0]=1;
	while(!Q.empty())
	{
		node f=Q.front();Q.pop();
		if(f.v1==c||f.v2==c)
		{
			cout<<f.op<<endl;
			return ;
		}
		if(f.v1!=m)
		{
			t.v1=m;
			t.op=f.op+1;
			t.v2=f.v2;
			if(!vis[t.v1][t.v2])
			{
			 vis[t.v1][t.v2]=1;
			 Q.push(t);
			}
		}
		if(f.v2!=n)
		{
			t.v2=n;
			t.op=f.op+1;
			t.v1=f.v1;
			if(!vis[t.v1][t.v2])
			{
			 vis[t.v1][t.v2]=1;
			 Q.push(t);
			}
		}
		if(f.v1!=0)
		{
			t.v1=0;
			t.v2=f.v2;
			t.op=f.op+1;
			if(!vis[t.v1][t.v2])
			{
			 vis[t.v1][t.v2]=1;
			 Q.push(t);
			}
		}
		if(f.v2!=0)
		{
			t.v2=0;
			t.v1=f.v1;
			t.op=f.op+1;
			if(!vis[t.v1][t.v2])
			{
			 vis[t.v1][t.v2]=1;
			 Q.push(t);
			}
		}
		if(f.v2!=0&&f.v1!=m)
		{
			t.v2=f.v2-(m-f.v1);if(t.v2<0) t.v2=0;
			t.v1=f.v1+f.v2;  if(t.v1>m) t.v1=m;
			t.op=f.op+1;
			if(!vis[t.v1][t.v2])
			{
			 vis[t.v1][t.v2]=1;
			 Q.push(t);
			}
		}
		if(f.v1!=0&&f.v2!=n)
		{
			t.v1=f.v1-(n-f.v2);if(t.v1<0) t.v1=0;
			t.v2=f.v2+f.v1;  if(t.v2>n) t.v2=n;
			t.op=f.op+1;
			if(!vis[t.v1][t.v2])
			{
			 vis[t.v1][t.v2]=1;
			 Q.push(t);
			}
		}
	}
	puts("impossible");
}
int main()
{
	
	while(cin>>m>>n>>c)
	{
		memset(vis,0,sizeof(vis));
		bfs();
	}
	return 0;
}


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