Mother's Milk

Problem：

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

 

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My Answer：

/* ID:Iby071 LANG:C++ TASK:milk3 */ #include<iostream> #include<fstream> using namespace std; static int a,b,c; static bool cfinal[21]; static bool occured[21][21]; int min(int x,int y) { return((x<y)?x:y); } void dfs(int s1,int s2,int s3) { if(occured[s1][s2]) return; //DFS终止条件：a，b，c当前状态已出现过 occured[s1][s2]=true; //标记已发生 if(s1==0) cfinal[s3]=true; //标记a里为0时的c状态 int delta=min(s1+s2,b)-s2; //六种倒牛奶的方式 if (delta>0) dfs(s1-delta,s2+delta,s3); delta=min(s1+s3,c)-s3; if (delta>0) dfs(s1-delta,s2,s3+delta); delta=min(s2+s1,a)-s1; if (delta>0) dfs(s1+delta,s2-delta,s3); delta=min(s2+s3,c)-s3; if (delta>0) dfs(s1,s2-delta,s3+delta); delta=min(s3+s1,a)-s1; if (delta>0) dfs(s1+delta,s2,s3-delta); delta=min(s3+s2,b)-s2; if (delta>0) dfs(s1,s2+delta,s3-delta); } int main() { ifstream fin("milk3.in"); ofstream fout("milk3.out"); int i; fin>>a>>b>>c; for(i=0;i<21;++i) cfinal[i]=false; for(i=0;i<21;++i) for(int j=0;j<21;++j) occured[i][j]=false; cfinal[c]=true; dfs(0,0,c); for(i=0;i<21;++i) if(cfinal[i]) {fout<<i;break;} for(++i ;i<21;++i) if(cfinal[i]) fout<<' '<<i; fout<<endl; return 0; }