Pots（POJ 3414）

最新推荐文章于 2021-12-12 20:12:21 发布

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本文介绍了一个算法，用于找到将两个容量为A和B升的水壶中的水移动到其中一个水壶中恰好为C升所需执行的最短操作序列。详细解释了如何通过FILL、DROP和POUR操作来实现这一目标，并提供了实现该算法的代码示例。

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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)

代码：

#include<stdio.h>
#include<iostream>
#include<string>
#include<string.h>
using namespace std;
#define MAXN 110
string str[7]={"","FILL(1)","FILL(2)","DROP(1)","DROP(2)","POUR(1,2)","POUR(2,1)"};
int a,b,c,l,r;
struct Info{
    int x;
    int y;
    int ope;
    int pre;
}info[MAXN*MAXN];
int visit[MAXN][MAXN],step[MAXN*MAXN];
void print(int r){
    int a;
    for(a=0;r;a++){
        step[a]=r;
        r=info[r].pre;
    }
    int index;
    cout<<a<<endl;
    for(a=a-1;a>=0;a--){
        r=step[a];
        index=info[r].ope;
        cout<<str[index]<<endl;
    }
}
bool solve(int tmpx,int tmpy,int ss){
    if(!visit[tmpx][tmpy]){
        info[r].x=tmpx;
        info[r].y=tmpy;
        info[r].ope=ss;
        info[r].pre=l;
        visit[tmpx][tmpy]=1;
        if(tmpx==c||tmpy==c){
            print(r);
            return true;
        }
        r++;
    }
    return false;
}
void BFS(){
    l=0;
    r=1;
    info[0].x=info[0].y=0;
    visit[0][0]=1;
    int tmpx,tmpy;
    while(l<r){
        //FILL(1) step1
        tmpx=a;
        tmpy=info[l].y;
        if(solve(tmpx,tmpy,1)) return;
        //FILL(2) step2
        tmpx=info[l].x;
        tmpy=b;
        if(solve(tmpx,tmpy,2)) return;
        //DROP(1) step3
        tmpx=0;
        tmpy=info[l].y;
        if(solve(tmpx,tmpy,3)) return;
        //DROP(2) step4
        tmpx=info[l].x;
        tmpy=0;
        if(solve(tmpx,tmpy,4)) return;
        //POUR(1,2) step5
        if((info[l].x+info[l].y)>=b){
            tmpx=info[l].x+info[l].y-b;
            tmpy=b;
        }else{
            tmpx=0;
            tmpy=info[l].x+info[l].y;
        }
        if(solve(tmpx,tmpy,5)) return;
        //POUR(2,1) step6
        if((info[l].x+info[l].y)>=a){
            tmpx=a;
            tmpy=info[l].x+info[l].y-a;
        }else{
            tmpx=info[l].x+info[l].y;
            tmpy=0;
        }
        if(solve(tmpx,tmpy,6)) return;
        l++;
    }
    cout<<"impossible"<<endl;
}
int main(){
    memset(visit,0,sizeof(visit));
    cin>>a>>b>>c;
    BFS();
    return 0;
}