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
1 /* 2 ID:fffgrdcc1 3 PROB:milk3 4 LANG:C++ 5 */ 6 #include<cstdio> 7 #include<iostream> 8 #include<algorithm> 9 using namespace std; 10 bool bo[21][21][21]={0}; 11 struct str 12 { 13 int a,b,c; 14 }e[10000]; 15 int cnt=1; 16 int q[10000],tail,head; 17 int a,b,c,A,B,C; 18 int main() 19 { 20 freopen("milk3.in","r",stdin); 21 freopen("milk3.out","w",stdout); 22 scanf("%d%d%d",&A,&B,&C); 23 head=-1;tail=0;e[0].a=e[0].b=0;e[0].c=C;q[0]=0;bo[0][0][C]=1; 24 while(head<tail) 25 { 26 head++; 27 int temp; 28 a=e[head].a,b=e[head].b,c=e[head].c; 29 temp=min(a,C-c);//a2c 30 a-=temp;c+=temp; 31 if(!bo[a][b][c]) 32 { 33 bo[a][b][c]=1; 34 q[++tail]=cnt; 35 e[++cnt].a=a; 36 e[cnt].b=b; 37 e[cnt].c=c; 38 } 39 a+=temp;c-=temp; 40 41 temp=min(A-a,c);//c2a 42 a+=temp;c-=temp; 43 if(!bo[a][b][c]) 44 { 45 bo[a][b][c]=1; 46 q[++tail]=cnt; 47 e[++cnt].a=a; 48 e[cnt].b=b; 49 e[cnt].c=c; 50 } 51 a-=temp;c+=temp; 52 53 temp=min(a,B-b);//a2b 54 a-=temp;b+=temp; 55 if(!bo[a][b][c]) 56 { 57 bo[a][b][c]=1; 58 q[++tail]=cnt; 59 e[++cnt].a=a; 60 e[cnt].b=b; 61 e[cnt].c=c; 62 } 63 a+=temp;b-=temp; 64 65 temp=min(A-a,b);//b2a 66 a+=temp;b-=temp; 67 if(!bo[a][b][c]) 68 { 69 bo[a][b][c]=1; 70 q[++tail]=cnt; 71 e[++cnt].a=a; 72 e[cnt].b=b; 73 e[cnt].c=c; 74 } 75 a-=temp;b+=temp; 76 77 temp=min(b,C-c);//b2c 78 b-=temp;c+=temp; 79 if(!bo[a][b][c]) 80 { 81 bo[a][b][c]=1; 82 q[++tail]=cnt; 83 e[++cnt].a=a; 84 e[cnt].b=b; 85 e[cnt].c=c; 86 } 87 b+=temp;c-=temp; 88 89 temp=min(c,B-b);//c2b 90 c-=temp;b+=temp; 91 if(!bo[a][b][c]) 92 { 93 bo[a][b][c]=1; 94 q[++tail]=cnt; 95 e[++cnt].a=a; 96 e[cnt].b=b; 97 e[cnt].c=c; 98 } 99 b-=temp;c+=temp; 100 } 101 int firflag=1; 102 for(int i=0;i<=C;i++) 103 { 104 b=C-i; 105 if(bo[0][b][i]) 106 if(firflag) 107 printf("%d",i),firflag=0; 108 else printf(" %d",i); 109 } 110 printf(" "); 111 return 0; 112 }
对了，输出格式很重要，提交前别忘记检查，血与泪的教训