Mother's Milk_usaco1.4_bfs

Mother’s Milk

Description

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.

PROGRAM NAME: milk3

INPUT FORMAT

A single line with the three integers A, B, and C.

OUTPUT FORMAT

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.

ANALYSIS

讲一下题目大意吧：

有三个桶ABC有各自的容量，Ｃ桶一开始装满了牛奶

请问经过若干次不流失不蒸发的倾倒后A桶为空的情况下C桶有哪些可能

这个专题好象就是搜索的，然后也似乎做过类似的题目

bfs寻找各种状态，有6种新途径：

1. a->b

2. a->c

3. b->c

4. b->a

5. c->b

6. c->a

枚举、判重、入队就好，这种题目做不出来简直了噜

输出最后不能有空格

CODE

/*
ID:wjp13241
PROG:milk3
LANG:C++
*/

#include <stdio.h>
#include <queue>
using namespace std;

struct milk
{
int a,b,c;
};

queue<milk>q;
priority_queue<int>ans;

bool f[21][21],t[21];

int max(int x,int y)
{
return x<y?y:x;
}

int min(int x,int y)
{
return x<y?x:y;
}

int main()
{
freopen("milk3.in","r",stdin);
freopen("milk3.out","w",stdout);

milk st;
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
st.a=0;
st.b=0;
st.c=c;
q.push(st);

while (q.size())
{
milk s,now=q.front();
q.pop();
if (now.a>=0&&now.b>=0&&now.c>=0)
{
f[now.a][now.b]=true;

if (!now.a&&!t[now.c])
{
ans.push(-now.c);
t[now.c]=true;
}

s=now;

now.a=max(0,now.a+now.b-b);
now.b=min(s.a+now.b,b);
if (!f[now.a][now.b])
q.push(now);

now=s;
now.a=max(0,now.a+now.c-c);
now.c=min(s.a+now.c,c);
if (!f[now.a][now.b])
q.push(now);

now=s;
now.b=max(0,now.b+now.a-a);
now.a=min(s.b+now.a,a);
if (!f[now.a][now.b])
q.push(now);

now=s;
now.b=max(0,now.b+now.c-c);
now.c=min(s.b+now.c,c);
if (!f[now.a][now.b])
q.push(now);

now=s;
now.c=max(0,now.c+now.b-b);
now.b=min(s.c+now.b,b);
if (!f[now.a][now.b])
q.push(now);

now=s;
now.c=max(0,now.c+now.a-a);
now.a=min(s.c+now.a,a);
if (!f[now.a][now.b])
q.push(now);
}
}

while (ans.size()>1)
{
printf("%d ",-ans.top());
ans.pop();
}
printf("%d\n",-ans.top());

fclose(stdin);
fclose(stdout);
return 0;
}