题目链接：https://cn.vjudge.net/problem/HDU-1495
题意：有一瓶Ｓ升的可乐，给容量分别为Ａ，Ｂ的两个杯子倒可乐，它们三个之间可以互相倒可乐。问这个三个容器之间的两个容器能不能平分Ｓ升的可乐。
思路：如果Ｓ是奇数的话，一定不能平分的.然后有六种倒法嘛，S-A,A-S,S-B,B-S,A-B,B-A。因为不清楚刻度，所以只有两种情况，１.一个容器里的可乐全倒进另一个容器 ２．一个容器将另一个容器倒满并且还剩可乐。将三个容器出现过的情况用一个三维数组进行标记。
注意：一开始标记的是vis[s][0][0]
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 #include <queue> 5 #include <stack> 6 #include <algorithm> 7 #include <cmath> 8 #include <map> 9 #define mem(a,b) memset(a,b,sizeof(a)); 10 using namespace std; 11 #define INF 0x3f3f3f3f 12 typedef long long ll; 13 int dir[4][2] = {0,1,0,-1,1,0,-1,0}; 14 const int maxn = 5000005; 15 int s,n,m,vis[105][105][105],ans; 16 bool flag; 17 struct node { 18 int x,y,z,w; 19 node(int x1,int y1,int z1,int w1):x(x1),y(y1),z(z1),w(w1){}; 20 }; 21 void BFS(int x1,int y1,int z1) { 22 queue<node>q; 23 q.push(node(x1,y1,z1,0)); 24 vis[x1][y1][z1] = 1; 25 int av = x1 / 2; 26 while(!q.empty()) { 27 node temp= q.front(); 28 q.pop(); 29 if(temp.x == av&&temp.y == av) { 30 flag = 1; 31 ans = temp.w; 32 break; 33 } 34 else if(temp.x == av&&temp.z == av) { 35 flag = 1; 36 ans = temp.w; 37 break; 38 } 39 else if(temp.z == av&&temp.y == av) { 40 flag = 1; 41 ans = temp.w; 42 break; 43 } 44 int fx,fy,fz; 45 //s-a 46 if(temp.x > (n - temp.y)) { 47 fy = n; 48 fx = temp.x+temp.y-n; 49 fz = temp.z; 50 } 51 else { 52 fy = temp.x + temp.y; 53 fx = 0; 54 fz = temp.z; 55 } 56 if(!vis[fx][fy][fz]) { 57 vis[fx][fy][fz] = 1; 58 q.push(node(fx,fy,fz,temp.w+1)); 59 } 60 //a-s 61 fy = 0; 62 fz = temp.z; 63 fx = temp.x + temp.y; 64 if(!vis[fx][fy][fz]) { 65 vis[fx][fy][fz] = 1; 66 q.push(node(fx,fy,fz,temp.w+1)); 67 } 68 //s-b 69 if(temp.x > (m - temp.z)) { 70 fy = temp.y; 71 fx = temp.x+temp.z-m; 72 fz = m; 73 } 74 else { 75 fy = temp.y; 76 fx = 0; 77 fz = temp.z + temp.x; 78 } 79 if(!vis[fx][fy][fz]) { 80 vis[fx][fy][fz] = 1; 81 q.push(node(fx,fy,fz,temp.w+1)); 82 } 83 //b-s 84 fx = temp.x + temp.z; 85 fy = temp.y; 86 fz = 0; 87 if(!vis[fx][fy][fz]) { 88 vis[fx][fy][fz] = 1; 89 q.push(node(fx,fy,fz,temp.w+1)); 90 } 91 //a-b 92 if(temp.y > (m- temp.z)) { 93 fz = m; 94 fy = temp.y + temp.z - m; 95 fx = temp.x; 96 } 97 else { 98 fy = 0; 99 fz = temp.y + temp.z; 100 fx = temp.x; 101 } 102 if(!vis[fx][fy][fz]) { 103 vis[fx][fy][fz] = 1; 104 q.push(node(fx,fy,fz,temp.w+1)); 105 } 106 //b-a 107 if(temp.z > (n - temp.y)) { 108 fy = n; 109 fx = temp.x; 110 fz = temp.z + temp.y - n; 111 } 112 else { 113 fz = 0; 114 fy = temp.z + temp.y; 115 fx = temp.x; 116 } 117 if(!vis[fx][fy][fz]) { 118 vis[fx][fy][fz] = 1; 119 q.push(node(fx,fy,fz,temp.w+1)); 120 } 121 122 } 123 } 124 int main() 125 { 126 while(cin >> s >> n >> m && s && m && n) { 127 mem(vis,0); 128 flag = 0; 129 if(s % 2 == 1) { 130 cout << "NO" << endl; 131 continue; 132 } 133 BFS(s,0,0); 134 if(flag) { 135 cout << ans << endl; 136 } 137 else { 138 cout << "NO" << endl; 139 } 140 } 141 return 0; 142 }