（BFS） 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:

1. FILL(i)        fill the pot i (1 ≤ i ≤ 2) from the tap;
2. DROP(i)      empty the pot i to the drain;
3. 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)

总共6种操作，为达到目标状态BFS一下即可。由于需要还原最短路径，在BFS的过程中记录一下达到每个状态前的状态，及进行操作的编号。倒序输出即可。

#include <iostream>
#include <string>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <vector>
#include <stack>
#define mp make_pair
//#define P make_pair
#define MIN(a,b) (a>b?b:a)
//#define MAX(a,b) (a>b?a:b)
typedef long long ll;
typedef unsigned long long ull;
const int MAX=1e5+5;
const int INF=1e8+5;
using namespace std;
//const int MOD=1e9+7;
typedef pair<ll,int> pii;
const double eps=0.00000001;

int A,B,C;
bool vi[105][105];
map<pii,pii>pre;
map<pii,int>id;
stack<int>st;
pii bfs()
{
    int a,b,a2,b2,num;
    vi[0][0]=true;
    queue<pii>que;
    que.push(mp(0,0));
    while(!que.empty())
    {
        pii tem=que.front(),tem2;
        que.pop();
        a=tem.first,b=tem.second;
        if(a==C||b==C)
            return tem;
        //fill 1
        if(!vi[A][b])
        {
            tem2=mp(A,b);
            pre[tem2]=tem;
            id[tem2]=1;
            que.push(tem2);
            vi[A][b]=true;
        }
        if(!vi[a][B])
        {
            tem2=mp(a,B);
            id[tem2]=2;
            pre[tem2]=tem;
            que.push(tem2);
            vi[a][B]=true;
        }
        if(!vi[0][b])
        {
            tem2=mp(0,b);
            id[tem2]=3;
            pre[tem2]=tem;
            que.push(tem2);
            vi[0][b]=true;
        }
        if(!vi[a][0])
        {
            tem2=mp(a,0);
            id[tem2]=4;
            pre[tem2]=tem;
            que.push(tem2);
            vi[a][0]=true;
        }
        num=B-b;
        b2=b+min(num,a);
        a2=a-min(num,a);
        if(!vi[a2][b2])
        {
            tem2=mp(a2,b2);
            id[tem2]=5;
            pre[tem2]=tem;
            que.push(tem2);
            vi[a2][b2]=true;
        }
        num=A-a;
        a2=a+min(num,b);
        b2=b-min(num,b);
        if(!vi[a2][b2])
        {
            tem2=mp(a2,b2);
            id[tem2]=6;
            pre[tem2]=tem;
            que.push(tem2);
            vi[a2][b2]=true;
        }
    }
    return mp(-1,-1);
}
int main()
{
    scanf("%d%d%d",&A,&B,&C);
    vi[0][0]=true;
    pii an=bfs();
    if(an.first==-1)
        return 0*printf("impossible
");
    while(!(an.first==0&&an.second==0))
    {
        st.push(id[an]);
        an=pre[an];
    }
    printf("%d
",st.size());
    while(!st.empty())
    {
        int oh=st.top();
        st.pop();
        if(oh<=2)
            printf("FILL(%d)
",oh);
        else if (oh<=4)
            printf("DROP(%d)
",oh-2);
        else if (oh==5)
            printf("POUR(1,2)
");
        else
            printf("POUR(2,1)
");
    }
}