1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #define sc(x) scanf("%d",&(x)) 6 #define pf(x) printf("%d ", x) 7 #define CL(x, y) memset(x, y, sizeof(x)) 8 using namespace std; 9 const int MAX = 105; 10 int front, rear; 11 int used[MAX][MAX]; 12 char str[6][10]= {"FILL(1)", "FILL(2)", "DROP(1)", "DROP(2)", "POUR(1,2)", "POUR(2,1)"}; 13 struct node 14 { 15 int a, b; 16 int n; 17 int step; 18 node *pre; 19 } Q[MAX*MAX]; //数组模拟队列 queue<Node> q; 20 void back(node a); 21 void BFS(int a, int b, int c); 22 int main() 23 { 24 int a,b,c; 25 CL(used, 0); 26 sc(a); 27 sc(b); 28 sc(c); 29 BFS(a, b, c); 30 pf(Q[rear].step); 31 back(Q[rear]); 32 } 33 void back(node a) 34 { 35 if(a.pre != NULL) 36 { 37 //printf("%s ", str[a.n]); 38 back(*(a.pre));//递归，找出初始操作 39 printf("%s ", str[a.n]); 40 } 41 return ; 42 } 43 void BFS(int a, int b, int c) 44 { 45 front = 0; 46 rear = 0; 47 Q[front].a = 0; 48 Q[front].b = 0; 49 Q[front].pre = NULL; 50 Q[front].step = 0; 51 used[0][0] = 1; 52 rear++; 53 while(front != rear) 54 { 55 node now = Q[front]; //{0, 0, NULL, 0} 56 used[now.a][now.b] = 1; 57 if(!used[a][now.b] && now.a!=a)//FILL(1) 将 A 加满 58 { 59 Q[rear].n = 0; //记录执行的操作 60 Q[rear].step = Q[front].step + 1; //步骤 +1 61 Q[rear].a = a; 62 Q[rear].b = now.b; 63 Q[rear].pre = &Q[front]; 64 used[a][now.b] = 1; 65 if(a==c || now.b==c) break; //完成操作 66 rear++; 67 } 68 if(!used[now.a][b] && now.b!=b)//FILL(2) 将 B 加满 69 { 70 Q[rear].n = 1; 71 Q[rear].step = Q[front].step+1; 72 Q[rear].a = now.a; 73 Q[rear].b = b; 74 Q[rear].pre = &Q[front]; 75 used[now.a][b] = 1; 76 if(now.a==c || b==c) break; 77 rear++; 78 } 79 if(!used[0][now.b] && now.a!=0)//DROP(1) 80 { 81 Q[rear].n = 2; 82 Q[rear].step = Q[front].step + 1; 83 Q[rear].a = 0; 84 Q[rear].b = now.b; 85 Q[rear].pre = &Q[front]; 86 used[0][now.b] = 1; 87 if(now.b == c) break;//A中没有水 88 rear++; 89 } 90 if(!used[now.a][0] && now.b!=0)//DROP(2) 91 { 92 Q[rear].n = 3; 93 Q[rear].step = Q[front].step + 1; 94 Q[rear].a = now.a; 95 Q[rear].b = 0; 96 Q[rear].pre = &Q[front]; 97 used[now.a][0] = 1; 98 if(now.a == c) break;//B中没有水 99 rear++; 100 } 101 int x = now.a < b-now.b ? now.a : b-now.b; 102 if(!used[now.a-x][now.b+x])//POUR(1,2) 103 { 104 Q[rear].n = 4; 105 Q[rear].step = Q[front].step+1; 106 Q[rear].a = now.a - x; 107 Q[rear].b = now.b + x; 108 Q[rear].pre = &Q[front]; 109 used[now.a-x][now.b+x] = 1; 110 if(now.a-x==c || now.b+x==c) break; 111 rear++; 112 } 113 int y = a-now.a < now.b ? a-now.a : now.b;//找到与自己本壶最接近的到水量 114 if(!used[now.a+y][now.b-y])//POUR(2,1) 115 { 116 Q[rear].n = 5; 117 Q[rear].step = Q[front].step+1; 118 Q[rear].a = now.a+y; 119 Q[rear].b = now.b-y; 120 Q[rear].pre = &Q[front]; 121 used[now.a+y][now.b-y] = 1; 122 if(now.a+y==c || now.b-y==c) break; 123 rear++; 124 } 125 front++; 126 }//跳出循环时， front==rear 127 if(front == rear) printf("impossible ");//条件判断 128 }
//BFS倒水问题,对于需要打印解得广搜题，必须保存搜索状态和状态的父亲指针，然后逆推，根据状态和状态之间的关系