http://acm.hdu.edu.cn/showproblem.php?pid=1495
题目就不说了， 说说思路！
倒可乐 无非有6种情况:
1. S 向 M 倒
2. S 向 N 倒
3. N 向 M 倒
4. N 向 S 倒
5. M 向 S 倒
6. M 向 N 倒
根据上述的六种情况来进行模拟， 每次模拟的结果都放进队列里面。
注意： 还要用到标记数组，防止数据重复进入队列导致爆栈。
1 #include<stdio.h> 2 #include<stdlib.h> 3 #include<math.h> 4 #include<queue> 5 #include<string.h> 6 #include<iostream> 7 using namespace std; 8 #define maxn 110 9 10 struct Node 11 { 12 int m;//M瓶子中装的可乐数m 13 int n; 14 int s; 15 int k;//总共倒的次数 16 }; 17 int M, N, S; 18 int BFS(Node P); 19 bool vis[maxn][maxn][maxn];//标记此情况是否有过, 防止重复 20 int main() 21 { 22 Node P; 23 while(scanf("%d%d%d",&S,&M,&N),M+N+S) 24 { 25 int t; 26 if(M < N)//将M存为最大的 27 t = M, M = N, N = t; 28 memset(vis,0,sizeof(vis)); 29 P.s = S, P.m = P.n = P.k = 0; 30 int ans = -1; 31 if(S%2 == 0)//奇数是没法平分的 32 ans = BFS(P); 33 34 if(ans == -1) 35 printf("NO "); 36 else 37 printf("%d ",ans); 38 } 39 return 0; 40 } 41 42 int BFS(Node P) 43 { 44 queue<Node> Q; 45 Q.push(P); 46 int x; 47 while( !Q.empty() ) 48 { 49 Node Pn; 50 Pn = Q.front(); 51 Q.pop(); 52 53 if(Pn.s == S/2 && Pn.m == S/2 )//当两个瓶子都是S/2则满足 54 return Pn.k; 55 56 P.k = Pn.k + 1; 57 /*下面就是倒可乐的过程*/ 58 if(Pn.s < S) 59 { 60 x = S - Pn.s; 61 62 if(Pn.n > x && !vis[S][Pn.n-x][Pn.m]) 63 { 64 vis[S][Pn.n-x][Pn.m] = 1; 65 P.s = S, P.m = Pn.m, P.n = Pn.n-x; 66 Q.push(P); 67 } 68 else if(Pn.n <= x && !vis[Pn.s+Pn.n][0][Pn.m]) 69 { 70 vis[Pn.s+Pn.n][0][Pn.m] = 1; 71 P.s = Pn.s + Pn.n, P.m = Pn.m, P.n = 0; 72 Q.push(P); 73 } 74 75 if(Pn.m > x && !vis[S][Pn.n][Pn.m-x]) 76 { 77 vis[S][Pn.n][Pn.m-x] = 1; 78 P.s = S, P.n = Pn.n, P.m = Pn.m-x; 79 Q.push(P); 80 } 81 else if(Pn.m <= x && !vis[Pn.s+Pn.m][Pn.n][0]) 82 { 83 vis[Pn.s+Pn.m][Pn.n][0] = 1; 84 P.s = Pn.s+Pn.m, P.n = Pn.n, P.m = 0; 85 Q.push(P); 86 } 87 } 88 89 if(Pn.n < N) 90 { 91 x = N - Pn.n; 92 93 if(Pn.s > x && !vis[Pn.s-x][N][Pn.m]) 94 { 95 vis[Pn.s-x][N][Pn.m] = 1; 96 P.s = Pn.s-x, P.n = N, P.m = Pn.m; 97 Q.push(P); 98 } 99 else if(Pn.s <= x && !vis[0][Pn.n+Pn.s][Pn.m]) 100 { 101 vis[0][Pn.n+Pn.s][Pn.m] = 1; 102 P.s = 0, P.n = Pn.n+Pn.s, P.m = Pn.m; 103 Q.push(P); 104 } 105 106 if(Pn.m > x && !vis[Pn.s][N][Pn.m-x]) 107 { 108 vis[Pn.s][N][Pn.m-x] = 1; 109 P.s = Pn.s, P.n = N, P.m = Pn.m-x; 110 Q.push(P); 111 } 112 else if(Pn.m <= x && !vis[Pn.s][Pn.n+Pn.m][0]) 113 { 114 vis[Pn.s][Pn.n+Pn.m][0] = 1; 115 P.s = Pn.s, P.n = Pn.n+Pn.m, P.m = 0; 116 Q.push(P); 117 } 118 } 119 120 if(Pn.m < M) 121 { 122 x = M - Pn.m; 123 124 if(Pn.s > x && !vis[Pn.s-x][Pn.n][M]) 125 { 126 vis[Pn.s-x][Pn.n][M] = 1; 127 P.s = Pn.s-x, P.n = Pn.n, P.m = M; 128 Q.push(P); 129 } 130 else if(Pn.s <= x && !vis[0][Pn.n][Pn.m+Pn.s]) 131 { 132 vis[0][Pn.n][Pn.m+Pn.s] = 1; 133 P.s = 0, P.n = Pn.n, P.m = Pn.m+Pn.s; 134 Q.push(P); 135 } 136 137 if(Pn.n > x && !vis[Pn.s][Pn.n-x][M]) 138 { 139 vis[Pn.s][Pn.n-x][M] = 1; 140 P.s = Pn.s, P.n = Pn.n-x, P.m = M; 141 Q.push(P); 142 } 143 else if(Pn.n <= x && !vis[Pn.s][0][Pn.m+Pn.n]) 144 { 145 vis[Pn.s][0][Pn.m+Pn.n] = 1; 146 P.s = Pn.s-x, P.n =0, P.m = Pn.m+Pn.n; 147 Q.push(P); 148 } 149 } 150 } 151 return -1; 152 }