POJ - 3414 Pots(BFS)

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Pots

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 27549		Accepted: 11558		Special Judge

Description

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.

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

Northeastern Europe 2002, Western Subregion

題意：

• 兩鍋1,2，他們分別裝有a,b升水。我們可以進行以下三個操作（其實是六個）

1. 把鍋i（i=1或2）加滿水。FILL(1)或FILL(2)
2. 把 鍋i 的水倒入 鍋j 中，如果j滿了，則不能再繼續倒水。POUR(1,2)或POUR(2,1)
3. 把鍋i 的水倒掉。DROP(1)或DROP(2)

• 一共六個操作，直接bfs就好了，代碼很清晰易懂的。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<set>
#include<algorithm>
#define sc(x) scanf("%d",&x)
#define scc(x,y) scanf("%d%d",&x,&y)
#define scll(x) scanf("%lld",&x)
#define mem(x,a) memset(x,a,sizeof x)
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;

const int maxn = 1005;

string op[]={"FILL(1)","FILL(2)","DROP(1)","DROP(2)","POUR(1,2)","POUR(2,1)"};
bool vis[maxn][maxn];

struct pot{
    int a,b;    //當前1,2中的水量
    vector<int> sq;    // 記錄路徑
};

void bfs(int a,int b,int c){
    vis[a][b]=1;    
    mem(vis,0);
    queue<pot> Q;
    pot t;
    t.a = t.b =0;
    t.sq.clear();
    vis[0][0]=1;
    Q.push(t);
    while(!Q.empty()){
        pot f = Q.front();
        Q.pop();
        if(f.a==c || f.b==c){
            cout<<f.sq.size()<<endl;
            for(int i=0;i<f.sq.size();i++){
                cout<<op[f.sq[i]]<<endl;
            }
            return;
        }
        pot tmp;

        for(int i=0;i<6;i++){
            if(i==0){                       // FILL(1)
                tmp.a=a;
                tmp.b=f.b;
            }
            if(i==1){   //FILL(2)
                tmp.b=b;
                tmp.a = f.a;
            }
            if(i==2){                       // DROP(1)
                tmp.a=0;
                tmp.b=f.b;
            }
            if(i==3){                       // DROP(2)
                tmp.a=f.a;
                tmp.b=0;
            }
            if(i==4){                       // POUR(1,2)
                if(f.a > b-f.b){
                    tmp.b = b;
                    tmp.a = f.a-(b-f.b);
                }
                else{
                    tmp.a=0;
                    tmp.b=f.a+f.b;
                }
            }
            if(i==5){                       // POUR(2,1)
                if(f.b > a-f.a){
                    tmp.a = a;
                    tmp.b = f.b-(a-f.a);
                }
                else{
                    tmp.b=0;
                    tmp.a=f.b+f.a;
                }
            }
            if(!vis[tmp.a][tmp.b]){         //沒有到達的狀態，更新入隊
                vis[tmp.a][tmp.b]=1;
                tmp.sq.assign(f.sq.begin(),f.sq.end());//把路徑賦值給下一個要進行入隊的狀態
                tmp.sq.push_back(i); // 把當前的操作路徑在放進當前要放進隊列狀態中
                Q.push(tmp);
            }
        }
    }
    printf("impossible\n");
}

int main(){
    int a,b,c;
    scanf("%d%d%d",&a,&b,&c);
    bfs(a,b,c);
}