整理自http://www.cnblogs.com/yzm10/p/7242590.html
- 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)
- 分析
- 让你求最少操作步数，每一步都有多种选择，所以用BFS，记录操作步骤。对于我来说实现起来比较难
- 1、要想到用01二维数组表示两个杯子的容量
- 2、想到有六种操作，1倒满，2倒满，倒空1……
- 3、操作状态Node的成员。
- 参考代码
#include<stdio.h> #include<queue> using namespace std; int b[105][105],re[10005];///标记数组和结果数组 struct Node{ int x,y,s,c,f,d; }node[10005]; int main() { int x,y,z,c,f,i;///f:操作成功与否的标志;c,i是数组遍历下标 queue<Node> q; scanf("%d%d%d",&x,&y,&z);///A,B,C if(z==0) printf("0 "); ///一定注意 else{ b[0][0]=1; node[1].x=0;///A node[1].y=0;///B node[1].s=0;///step步骤，统计需要几步 node[1].c=1; node[1].f=1;///记下上一步的C，以便找出当前操作的上一步操作 node[1].d=0;///6类操作的哪一类 q.push(node[1]); c=1;f=0; while(q.size()){ for(i=1;i<=6;i++){ if(i==1&&q.front().x<x&&b[x][q.front().y]==0){ b[x][q.front().y]=1;///倒满1 node[++c].x=x; node[c].y=q.front().y; node[c].s=q.front().s+1; node[c].c=c; node[c].f=q.front().c; node[c].d=1; if(node[c].x==z||node[c].y==z){ f=node[c].s; break; } q.push(node[c]); } else if(i==2&&q.front().y<y&&b[q.front().x][y]==0){ b[q.front().x][y]=1;///倒满2 node[++c].x=q.front().x; node[c].y=y; node[c].s=q.front().s+1; node[c].c=c; node[c].f=q.front().c; node[c].d=2; if(node[c].x==z||node[c].y==z){ f=node[c].s; break; } q.push(node[c]); } else if(i==3&&q.front().x>0&&b[0][q.front().y]==0){ b[0][q.front().y]=1;///倒空1 node[++c].x=0; node[c].y=q.front().y; node[c].s=q.front().s+1; node[c].c=c; node[c].f=q.front().c; node[c].d=3; if(node[c].x==z||node[c].y==z){ f=node[c].s; break; } q.push(node[c]); } else if(i==4&&q.front().y>0&&b[q.front().x][0]==0){ b[q.front().x][0]=1;///倒空2 node[++c].x=q.front().x; node[c].y=0; node[c].s=q.front().s+1; node[c].c=c; node[c].f=q.front().c; node[c].d=4; if(node[c].x==z||node[c].y==z){ f=node[c].s; break; } q.push(node[c]); } else if(i==5&&q.front().x>0&&q.front().y<y){ ///2满或不满的情况 int tx=q.front().x<y-q.front().y?0:q.front().x+q.front().y-y; int ty=q.front().x<y-q.front().y?q.front().x+q.front().y:y; if(b[tx][ty]==0){ b[tx][ty]=1; node[++c].x=tx; node[c].y=ty; node[c].s=q.front().s+1; node[c].c=c; node[c].f=q.front().c; node[c].d=5; if(node[c].x==z||node[c].y==z){ f=node[c].s; break; } q.push(node[c]); } } else if(i==6&&q.front().x<x&&q.front().y>0){ int tx=x-q.front().x<q.front().y?x:q.front().x+q.front().y; int ty=x-q.front().x<q.front().y?q.front().x+q.front().y-x:0; if(b[tx][ty]==0){ b[tx][ty]=1; node[++c].x=tx; node[c].y=ty; node[c].s=q.front().s+1; node[c].c=c; node[c].f=q.front().c; node[c].d=6; if(node[c].x==z||node[c].y==z){ f=node[c].s; break; } q.push(node[c]); } } } if(f!=0) break; q.pop(); } if(f==0) printf("impossible "); else{ printf("%d ",f); //根据node中f的标记找到最少的操作 for(i=1;i<=f;i++){ re[i]=node[c].d; c=node[c].f; } for(i=f;i>=1;i--){ if(re[i]==1) printf("FILL(1) "); else if(re[i]==2) printf("FILL(2) "); else if(re[i]==3) printf("DROP(1) "); else if(re[i]==4) printf("DROP(2) "); else if(re[i]==5) printf("POUR(1,2) "); else if(re[i]==6) printf("POUR(2,1) "); } } } return 0; }
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