USACO 1.4 Mother's Milk
Mother's Milk
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.
SAMPLE INPUT (file milk3.in)
8 9 10
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.
SAMPLE OUTPUT (file milk3.out)
1 2 8 9 10
SAMPLE INPUT (file milk3.in)
2 5 10
SAMPLE OUTPUT (file milk3.out)
5 6 7 8 9 10 题目大意:倒牛奶。。。。你有三个筒子ABC会告诉你容积,开始的时候AB都是空的,C是满的,问你在不把牛奶倒出三个筒子之外的情况下,在A桶是空的情况下,C桶有多少奶,顺序输出所有可能性。
思路:没什么说的了,BFS。代码写在下面
/*
ID:fffgrdcc1
PROB:milk3
LANG:C++
*/
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
bool bo[][][]={};
struct str
{
int a,b,c;
}e[];
int cnt=;
int q[],tail,head;
int a,b,c,A,B,C;
int main()
{
freopen("milk3.in","r",stdin);
freopen("milk3.out","w",stdout);
scanf("%d%d%d",&A,&B,&C);
head=-;tail=;e[].a=e[].b=;e[].c=C;q[]=;bo[][][C]=;
while(head<tail)
{
head++;
int temp;
a=e[head].a,b=e[head].b,c=e[head].c;
temp=min(a,C-c);//a2c
a-=temp;c+=temp;
if(!bo[a][b][c])
{
bo[a][b][c]=;
q[++tail]=cnt;
e[++cnt].a=a;
e[cnt].b=b;
e[cnt].c=c;
}
a+=temp;c-=temp; temp=min(A-a,c);//c2a
a+=temp;c-=temp;
if(!bo[a][b][c])
{
bo[a][b][c]=;
q[++tail]=cnt;
e[++cnt].a=a;
e[cnt].b=b;
e[cnt].c=c;
}
a-=temp;c+=temp; temp=min(a,B-b);//a2b
a-=temp;b+=temp;
if(!bo[a][b][c])
{
bo[a][b][c]=;
q[++tail]=cnt;
e[++cnt].a=a;
e[cnt].b=b;
e[cnt].c=c;
}
a+=temp;b-=temp; temp=min(A-a,b);//b2a
a+=temp;b-=temp;
if(!bo[a][b][c])
{
bo[a][b][c]=;
q[++tail]=cnt;
e[++cnt].a=a;
e[cnt].b=b;
e[cnt].c=c;
}
a-=temp;b+=temp; temp=min(b,C-c);//b2c
b-=temp;c+=temp;
if(!bo[a][b][c])
{
bo[a][b][c]=;
q[++tail]=cnt;
e[++cnt].a=a;
e[cnt].b=b;
e[cnt].c=c;
}
b+=temp;c-=temp; temp=min(c,B-b);//c2b
c-=temp;b+=temp;
if(!bo[a][b][c])
{
bo[a][b][c]=;
q[++tail]=cnt;
e[++cnt].a=a;
e[cnt].b=b;
e[cnt].c=c;
}
b-=temp;c+=temp;
}
int firflag=;
for(int i=;i<=C;i++)
{
b=C-i;
if(bo[][b][i])
if(firflag)
printf("%d",i),firflag=;
else printf(" %d",i);
}
printf("\n");
return ;
}
对了,输出格式很重要,提交前别忘记检查,血与泪的教训
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