Pots --bfs

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)

解题报告

bfs的搜索永远优先最小操作数的，所以再用dp[][]记录防止重复即可在o(A*B)找到答案。

为了方便回溯路径，用一个back[][]记录前驱即可

//Accepted 760kB 0ms
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
#define MAX_N 102
using namespace std;
struct node{int last_A,last_B,op;};
int dp[MAX_N][MAX_N];
node back[MAX_N][MAX_N];
int A,B,goal;
char str[7][10]={"nothing","FILL(1)","FILL(2)","DROP(2)","DROP(1)","POUR(1,2)","POUR(2,1)"};

void print(int n_A,int n_B){
    node e=back[n_A][n_B];
    if(e.last_A||e.last_B)
        print(e.last_A,e.last_B);
    puts(str[e.op]);
}

void bfs(){
    queue<pair<int,int> >que;//first-> A  second-> B
    memset(dp,-1,sizeof(dp));
    que.push(make_pair(0,0));
    dp[0][0]=0;//dp record min steps from start

    while(!que.empty()){
        int n_A=que.front().first,n_B=que.front().second;que.pop();
        int s=dp[n_A][n_B];
        if(n_A==goal||n_B==goal){//back
            printf("%d\n",s);
            print(n_A,n_B);
            return ;
        }

        //操作一步有六种情况
        if(dp[A][n_B]==-1){//FILL(1)
            dp[A][n_B]=s+1;
            back[A][n_B]={n_A,n_B,1};
            que.push(make_pair(A,n_B));
        }
        if(dp[n_A][B]==-1){//FILL(2)
            dp[n_A][B]=s+1;
            back[n_A][B]={n_A,n_B,2};
            que.push(make_pair(n_A,B));
        }

        if(dp[n_A][0]==-1){//DROP(2)
            dp[n_A][0]=s+1;
            back[n_A][0]={n_A,n_B,3};
            que.push(make_pair(n_A,0));
        }
        if(dp[0][n_B]==-1){//DROP(1)
            dp[0][n_B]=s+1;
            back[0][n_B]={n_A,n_B,4};
            que.push(make_pair(0,n_B));
        }

        int tmpb=n_A+n_B>B?B:n_A+n_B;
        int tmpa=n_A+n_B-tmpb;
        if(dp[tmpa][tmpb]==-1){//POUR(1,2)
            dp[tmpa][tmpb]=s+1;
            back[tmpa][tmpb]={n_A,n_B,5};
            que.push(make_pair(tmpa,tmpb));
        }
        tmpa=n_A+n_B>A?A:n_A+n_B;
        tmpb=n_A+n_B-tmpa;
        if(dp[tmpa][tmpb]==-1){//POUR(2,1)
            dp[tmpa][tmpb]=s+1;
            back[tmpa][tmpb]={n_A,n_B,6};
            que.push(make_pair(tmpa,tmpb));
        }
    }
    puts("impossible");
}

int main()
{
    while(~scanf("%d%d%d",&A,&B,&goal)){
        bfs();
    }
    return 0;
}