Pots
- 总时间限制:
- 1000ms
- 内存限制:
- 65536kB
- 描述
- You are given two pots, having the volume of A and B liters respectively. The following operations can be performed:
- FILL(i) fill the pot i (1 ≤ i ≤ 2) from the tap;
- DROP(i) empty the pot i to the drain;
- POUR(i,j) pour from pot i to pot j; after this operation either the pot j is full (and there may be some water left in the pot i), or the pot i is empty (and all its contents have been moved to the pot j).
Write a program to find the shortest possible sequence of these operations that will yield exactly C liters of water in one of the pots.
- 输入
- On the first and only line are the numbers A, B, and C. These are all integers in the range from 1 to 100 and C≤max(A,B).
- 输出
- The first line of the output must contain the length of the sequence of operations K. The following K lines must each describe one operation. If there are several sequences of minimal length, output any one of them. If the desired result can’t be achieved, the first and only line of the file must contain the word ‘impossible’.
- 样例输入
-
3 5 4
- 样例输出
-
6 FILL(2) POUR(2,1) DROP(1) POUR(2,1) FILL(2) POUR(2,1)
本来我是不会做的，看了这篇题解 http://blog.csdn.net/kindlucy/article/details/5827794 之后明白的……
还是BFS
设(i,j)为瓶1与瓶2在某一时刻的容量，那么从这点出发，可以到达的点有：
(A, j) : FILL(1)
(i, B) : FILL(2)
(0, j): DROP(1)
(i, 0): DROP(2)
(i+j, 0) or (A, j-A+i) : POUR(2,1)
(0, i+j) or (i-B+j, B) : POUR(1,2)
判断这些点的访问状态，然后广搜下去，另外，要求出从源顶点到这些点的最短路径
当i=C 或 j=C时，就找到了
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 | #include<iostream> #include<cstdio> #include<cstring> #include<queue> #include<stack> using namespace std; /* Starting From (0,0),End With (C,X) Or (X,C) Operations: FILL(1):(A,XB) FILL(2):(XA,B) DROP(1):(0,XB) DROP(2):(XA,0) POUR(2,1): POUR(1,2): */ struct State{ char C;//F:Fill;P:Pour;D:Drop int N1; int N2; int Parent_X; int Parent_Y; }; int A,B,C; int pos_x,pos_y; State S[101][101]; bool visited[101][101]; inline void Fill_State(int x,int y,char type,int N1,int N2,int Parent_X,int Parent_Y){ if(visited[x][y])return; S[x][y].C=type; S[x][y].N1=N1; S[x][y].N2=N2; S[x][y].Parent_X=Parent_X; S[x][y].Parent_Y=Parent_Y; } bool BFS(){ queue<int> N1,N2; N1.push(0); N2.push(0); while(!N1.empty()){ int t1=N1.front(),t2=N2.front(); N1.pop();N2.pop(); if(visited[t1][t2])continue; visited[t1][t2]=true; if(t1==C||t2==C){ pos_x=t1; pos_y=t2; return true; } //Operations if(t1!=A){//FIll A N1.push(A); N2.push(t2); Fill_State(A,t2,'F',1,0,t1,t2); } if(t2!=B){//FILL B N1.push(t1); N2.push(B); Fill_State(t1,B,'F',2,0,t1,t2); } if(t1!=0){//DROP A N1.push(0); N2.push(t2); Fill_State(0,t2,'D',1,0,t1,t2); } if(t2!=0){//DROP B N1.push(t1); N2.push(0); Fill_State(t1,0,'D',2,0,t1,t2); } if(t1!=0&&t2!=B){//POUR(A,B) if(t1<=B-t2){ N1.push(0); N2.push(t2+t1); Fill_State(0,t2+t1,'P',1,2,t1,t2); }else{ N1.push(t1-B+t2); N2.push(B); Fill_State(t1-B+t2,B,'P',1,2,t1,t2); } } if(t1!=A&&t2!=0){//POUR(B,A) if(t2<=A-t1){ N1.push(t2+t1); N2.push(0); Fill_State(t2+t1,0,'P',2,1,t1,t2); }else{ N1.push(A); N2.push(t2-A+t1); Fill_State(A,t2-A+t1,'P',2,1,t1,t2); } } } return false; } int main(){ scanf("%d%d%d",&A,&B,&C); memset(visited,false,sizeof(visited)); S[0][0].Parent_X=-1; S[0][0].Parent_Y=-1; bool flag=BFS(); if(!flag){ printf("impossible"); }else{ stack<State *> Stack_ins; for(State * i=&S[pos_x][pos_y];!(i->Parent_X==-1&&i->Parent_Y==-1);i=&S[i->Parent_X][i->Parent_Y])Stack_ins.push(i); printf("%d\n",Stack_ins.size()); while(!Stack_ins.empty()){ switch(Stack_ins.top()->C){ case 'F': printf("FILL(%d)\n",Stack_ins.top()->N1); break; case 'P': printf("POUR(%d,%d)\n",Stack_ins.top()->N1,Stack_ins.top()->N2); break; case 'D': printf("DROP(%d)\n",Stack_ins.top()->N1); break; } Stack_ins.pop(); } } } |