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Farmer John has three milking buckets of capacity A, B, and C liters. Each of the numbers A, B, and C is an integer from 1 through 20, inclusive. Initially, buckets A and B are empty while bucket C is full of milk. Sometimes, FJ pours milk from one bucket to another until the second bucket is filled or the first bucket is empty. Once begun, a pour must be completed, of course. Being thrifty, no milk may be tossed out.
Write a program to help FJ determine what amounts of milk he can leave in bucket C when he begins with three buckets as above, pours milk among the buckets for a while, and then notes that bucket A is empty.
A single line with the three integers A, B, and C.
8 9 10
A single line with a sorted list of all the possible amounts of milk that can be in bucket C when bucket A is empty.
1 2 8 9 10
2 5 10
5 6 7 8 9 10
题意： Farmer John有三个桶A,B,C，输入三个桶的容积;开始时桶A和第桶B为空，桶C为满。
Farmer John要做pours milk操作，把桶x的牛奶导入桶y中，除非桶y被倒满或桶x已倒空，一次操作算作结束。
进行n次倒牛奶操作改变C桶中牛奶的余量，但要求终止时A桶要为空。
问：C桶中牛奶的余量一共有多少种情况。
题解： 一共有6种操作 A->B;A->C; B->A;B->C; C->A;C->B;
用宽搜就可以了做出来，但感觉这次我写的代码好丑，过于冗余。
ps:本人大三狗一枚，正在持续更新博客，文章里有任何问题，希望各位网友可以指出。若有疑问也可在评论区留言，我会尽快回复。希望能与各位网友互相学习，谢谢
/*
ID: cxq_xia1
PROG: milk3
LANG: C++
*/
#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;
struct node
{
int a,b,c;
friend bool operator < (node x,node y)
{
return x.c > y.c;
}
};
bool flag[21][21][21];
queue<node> q;
priority_queue<node> ans;
int A,B,C,cnt=0;
node tmp1,tmp2;
void BFS()
{
tmp1.a=0;tmp1.b=0;tmp1.c=C;
flag[tmp1.a][tmp1.b][tmp1.c]=true;
q.push(tmp1);
while(!q.empty())
{
tmp1=q.front();
q.pop();
if(tmp1.a==0)
{
ans.push(tmp1);
cnt++;
}
if(tmp1.a!=0)
{
if(tmp1.b+tmp1.a<=B)
{
tmp2.a=0;
tmp2.b=tmp1.b+tmp1.a;
tmp2.c=tmp1.c;
}
else
{
tmp2.a=tmp1.a-(B-tmp1.b);
tmp2.b=B;
tmp2.c=tmp1.c;
}
if(!flag[tmp2.a][tmp2.b][tmp2.c])
{
flag[tmp2.a][tmp2.b][tmp2.c]=true;
q.push(tmp2);
}
if(tmp1.c+tmp1.a<=C)
{
tmp2.a=0;
tmp2.b=tmp1.b;
tmp2.c=tmp1.c+tmp1.a;
}
else
{
tmp2.a=tmp1.a-(C-tmp1.c);
tmp2.b=tmp1.b;
tmp2.c=C;
}
if(!flag[tmp2.a][tmp2.b][tmp2.c])
{
flag[tmp2.a][tmp2.b][tmp2.c]=true;
q.push(tmp2);
}
}
if(tmp1.b!=0)
{
if(tmp1.a+tmp1.b<=A)
{
tmp2.a=tmp1.a+tmp1.b;
tmp2.b=0;
tmp2.c=tmp1.c;
}
else
{
tmp2.a=A;
tmp2.b=tmp1.b-(A-tmp1.a);
tmp2.c=tmp1.c;
}
if(!flag[tmp2.a][tmp2.b][tmp2.c])
{
flag[tmp2.a][tmp2.b][tmp2.c]=true;
q.push(tmp2);
}
if(tmp1.c+tmp1.b<=C)
{
tmp2.a=tmp1.a;
tmp2.b=0;
tmp2.c=tmp1.c+tmp1.b;
}
else
{
tmp2.a=tmp1.a;
tmp2.b=tmp1.b-(C-tmp1.c);
tmp2.c=C;
}
if(!flag[tmp2.a][tmp2.b][tmp2.c])
{
flag[tmp2.a][tmp2.b][tmp2.c]=true;
q.push(tmp2);
}
}
if(tmp1.c!=0)
{
if(tmp1.a+tmp1.c<=A)
{
tmp2.a=tmp1.a+tmp1.c;
tmp2.b=tmp1.b;
tmp2.c=0;
}
else
{
tmp2.a=A;
tmp2.b=tmp1.b;
tmp2.c=tmp1.c-(A-tmp1.a);
}
if(!flag[tmp2.a][tmp2.b][tmp2.c])
{
flag[tmp2.a][tmp2.b][tmp2.c]=true;
q.push(tmp2);
}
if(tmp1.b+tmp1.c<=B)
{
tmp2.a=tmp1.a;
tmp2.b=tmp1.b+tmp1.c;
tmp2.c=0;
}
else
{
tmp2.a=tmp1.a;
tmp2.b=B;
tmp2.c=tmp1.c-(B-tmp1.b);
}
if(!flag[tmp2.a][tmp2.b][tmp2.c])
{
flag[tmp2.a][tmp2.b][tmp2.c]=true;
q.push(tmp2);
}
}
}
}
int main()
{
freopen("milk3.in","r",stdin);
freopen("milk3.out","w",stdout);
memset(flag,false,sizeof(flag));
cin >> A >> B >> C;
BFS();
node tmp;
int i;
while(!ans.empty())
{
tmp=ans.top();
ans.pop();
i++;
if(i!=cnt)
cout << tmp.c << " ";
else
cout << tmp.c <<endl ;
}
return 0;
}
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原文地址：http://www.cnblogs.com/WillsCheng/p/4796505.html
