| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 21332 | Accepted: 9105 | Special Judge | ||
Description
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.
Input
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).
Output
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’.
Sample Input
3 5 4
Sample Output
6 FILL(2) POUR(2,1) DROP(1) POUR(2,1) FILL(2) POUR(2,1)
Source
POUR（1,2），把杯子1的水倒入杯子2，POUR（2,1）把杯子2的水倒入1.
vis[i][j],表示1杯水量为i,2杯水量为j的情况是否已经出现过了，避免重复计算。。直接看代码吧
#include "iostream"
#include "queue"
#include "string"
#include "map"
using namespace std;
struct P//用结构体记录1,2杯子里面的水量和操作路径
{
int x,y;
string path;
};
map<int,string> mp;
bool vis[105][105];
int x,y,z,flag;
void init()//初始话，用map来记录对应关系
{
mp.clear();
mp[1]="FILL(1)";
mp[2]="FILL(2)";
mp[3]="DROP(1)";
mp[4]="DROP(2)";
mp[5]="POUR(1,2)";
mp[6]="POUR(2,1)";
}
P work(int t,P s)//计算出在这六中情况下1,2，杯里面的水量和路径
{
P a;
switch(t){
case 1:
a.x=x,a.y=s.y,a.path=s.path+'1';
break;
case 2:
a.x=s.x,a.y=y,a.path=s.path+'2';
break;
case 3:
a.x=0,a.y=s.y,a.path=s.path+'3';
break;
case 4:
a.x=s.x,a.y=0,a.path=s.path+'4';
break;
case 5:
if(y-s.y<s.x){
a.x=s.x-(y-s.y);
a.y=y;
}
else{
a.x=0;
a.y=s.x+s.y;
}
a.path=s.path+'5';
break;
case 6:
if(x-s.x<s.y){
a.y=s.y-(x-s.x);
a.x=x;
}
else{
a.x=s.x+s.y;
a.y= 0;
}
a.path=s.path+'6';
break;
}
return a;
}
void bfs()
{
queue<P> que;
P a;
a.x=x,a.y=0,a.path="1";//首次操作，将杯1的水加满或者将杯2的水加满
vis[x][0]=1;
que.push(a);
a.x=0,a.y=y,a.path="2";
vis[0][y]=1;
que.push(a);
while(!que.empty()){
P s=que.front();
que.pop();
if(s.x==z||s.y==z){
flag=1;
cout<<s.path.size()<<endl;
for(int i=0;i<s.path.size();i++)
cout<<mp[s.path[i]-'0']<<endl;
return;
}
for(int i=1;i<=6;i++){//六种情况
P a=work(i,s);
if(!vis[a.x][a.y]){
vis[a.x][a.y]=1;
que.push(a);
}
}
}
}
int main()
{
ios::sync_with_stdio(false);
cin>>x>>y>>z;
flag=0;
init();
bfs();
if(!flag)
cout<<"impossible"<<endl;
return 0;
}