(简单) POJ 3414 Pots，BFS+记录路径。

　　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.

　　就是对两个杯子进行操作了，装满，倒空，倒过去。两个水杯里的水表示一个状态，然后BFS就好，要记录路径的话，记住每一个的父亲，然后最后反着来一遍就好了。

代码如下：

#include<iostream>
#include<cstring>
#include<cmath>
#include<queue>

using namespace std;

int A,B,C;
int rem[][];
int shorem[];

struct state
{
    int a,b;
    int type;
    int faa,fab;
    int num;

    state() {}
};

state sta[][];

void showans(state *e)
{
    int cou=;

    while(e->a||e->b)
    {
        shorem[cou++]=e->type;

        e=&sta[e->faa][e->fab];
    }

    cout<<cou<<endl;
    for(int i=cou-;i>=;--i)
        switch(shorem[i])
        {
            case :
                cout<<"FILL(1)\n";
                break;
            case :
                cout<<"FILL(2)\n";
                break;
            case :
                cout<<"DROP(1)\n";
                break;
            case :
                cout<<"DROP(2)\n";
                break;
            case :
                cout<<"POUR(1,2)\n";
                break;
            case :
                cout<<"POUR(2,1)\n";
                break;
        }

}

void slove()
{
    queue <state *> que;
    state *temp;
    int t1,t2,t3;

    sta[][].faa=sta[][].fab=-;
    sta[][].type=;
    sta[][].num=;
    que.push(&sta[][]);

    while(!que.empty())
    {
        temp=que.front();
        que.pop();

        if(temp->a==C||temp->b==C)
        {
            showans(temp);
            return;
        }

        t1=temp->a;
        t2=temp->b;

        if(sta[A][t2].num==-)
        {
            sta[A][t2].num=temp->num+;
            sta[A][t2].faa=t1;
            sta[A][t2].fab=t2;
            sta[A][t2].type=;

            que.push(&sta[A][t2]);
        }
        if(sta[t1][B].num==-)
        {
            sta[t1][B].num=temp->num+;
            sta[t1][B].faa=t1;
            sta[t1][B].fab=t2;
            sta[t1][B].type=;

            que.push(&sta[t1][B]);
        }
        if(sta[][t2].num==-)
        {
            sta[][t2].num=temp->num+;
            sta[][t2].faa=t1;
            sta[][t2].fab=t2;
            sta[][t2].type=;

            que.push(&sta[][t2]);
        }
        if(sta[t1][].num==-)
        {
            sta[t1][].num=temp->num+;
            sta[t1][].faa=t1;
            sta[t1][].fab=t2;
            sta[t1][].type=;

            que.push(&sta[t1][]);
        }
        t3=min(t1,B-t2);
        if(sta[t1-t3][t2+t3].num==-)
        {
            sta[t1-t3][t2+t3].num=temp->num+;
            sta[t1-t3][t2+t3].faa=t1;
            sta[t1-t3][t2+t3].fab=t2;
            sta[t1-t3][t2+t3].type=;

            que.push(&sta[t1-t3][t2+t3]);
        }
        t3=min(t2,A-t1);
        if(sta[t1+t3][t2-t3].num==-)
        {
            sta[t1+t3][t2-t3].num=temp->num+;
            sta[t1+t3][t2-t3].faa=t1;
            sta[t1+t3][t2-t3].fab=t2;
            sta[t1+t3][t2-t3].type=;

            que.push(&sta[t1+t3][t2-t3]);
        }
    }

    cout<<"impossible\n";
}

int main()
{
    ios::sync_with_stdio(false);

    for(int i=;i<=;++i)
        for(int j=;j<=;++j)
        {
            sta[i][j].a=i;
            sta[i][j].b=j;
        }

    while(cin>>A>>B>>C)
    {
        for(int i=;i<=;++i)
            for(int j=;j<=;++j)
                sta[i][j].num=-;

        slove();
    }

    return ;
}