poj 3414

题意：给出两个容积分别为 a 和 b 的pot，按照以下三种操作方式，求出能否在一定步数后，使者两个pot的其中一个的水量为c。

      1.FILL(i)：将ipot倒满水。

      2.DROP(i)：将ipot倒空水。

      3.POUR(i,j)： 将ipot的水倒到jpot上，直至要么ipot为空，要么jpot为满。

典型的最短求解问题 并记录路径
#include<iostream>
#include<queue>
using namespace std;
bool v[111][111];
struct No
{
    int a,b;
    int step;
    int prex,prey;
};
struct Node
{
    int step;
    int prex,prey;
    char a[15];
}map[111][111];
int a,b,c;
void out(int x,int y)
{
    if(map[x][y].prex==-1&&map[x][y].prey==-1)
        return ;
    out(map[x][y].prex,map[x][y].prey);
    cout<<map[x][y].a<<endl;
}
bool bfs()
{
    queue<No>q;
    No now,next;
    now.a=0;now.b=0;
    now.step=0;
    now.prex=-1;now.prey=-1;
    memset(v,false,sizeof(v));
    v[0][0]=true;
    q.push(now);
    
    map[0][0].prex=-1;map[0][0].prey=-1;
    map[0][0].step=0;
    
    while(!q.empty())
    {
        now=q.front();
        if(now.a==c||now.b==c)
        {
            cout<<now.step<<endl;
            out(now.a,now.b);
            return true;
        }
        q.pop();
        
        
        //1填满
        if(now.a<a&&(!v[a][now.b]||map[a][now.b].step>now.step+1))
        {
            v[a][now.b]=1;
            next=now;
            next.a=a;
            next.step++;
            
            next.prex=now.a;
            next.prey=now.b;
            char s[]="FILL(1)";
            strcpy(map[next.a][next.b].a,s);
            map[next.a][next.b].step=now.step+1;
            map[next.a][next.b].prex=now.a;
            map[next.a][next.b].prey=now.b;
            v[a][next.b]=true;

            q.push(next);
        }
        if(now.b<b&&(!v[now.a][b]||map[now.a][b].step>now.step+1))
        {
            v[now.a][b]=1;
            next=now;
            next.b=b;
            next.step++;
            
            next.prex=now.a;
            next.prey=now.b;
            
            strcpy(map[next.a][next.b].a,"FILL(2)");
            map[next.a][next.b].step=now.step+1;
            map[next.a][next.b].prex=now.a;
            map[next.a][next.b].prey=now.b;
            v[next.a][next.b]=true;
            q.push(next);
        }
        // 清空
        if(now.a!=0&&(!v[0][now.b]||map[0][now.b].step>now.step+1))
        {
            v[0][now.b]=1;
            
            next=now;
            next.a=0;
            next.prex=now.a;
            next.prey=now.b;
            
            next.step++;
            
            strcpy(map[next.a][next.b].a,"DROP(1)");
            map[next.a][next.b].step=now.step+1;
            map[next.a][next.b].prex=now.a;
            map[next.a][next.b].prey=now.b;
            
            q.push(next);
        }
        if(now.b!=0&&(!v[now.a][0]||map[now.a][0].step>now.step+1))
        {
            v[now.a][0]=1;
            
            next=now;
            next.b=0;
            next.prex=now.a;
            next.prey=now.b;
            
            next.step++;
            
            strcpy(map[next.a][next.b].a,"DROP(2)");
            map[next.a][next.b].step=now.step+1;
            map[next.a][next.b].prex=now.a;
            map[next.a][next.b].prey=now.b;
            
            q.push(next);
        }
        //从i到到j
        if(now.a!=0)
        {
            int aa,bb;
            int x=b-now.b;//需要到得量
            if(x>=now.a)
            {
                aa=0;bb=now.b+now.a;
            }
            else
            {
                bb=b;aa=now.a-x;
            }
            if((!v[aa][bb]||map[aa][bb].step>now.step+1))
            {
                next.a=aa;next.b=bb;
                next.prex=now.a;next.prey=now.b;
                next.step=now.step+1;
                
                map[aa][bb].step=now.step+1;
                strcpy(map[aa][bb].a,"POUR(1,2)");
                map[next.a][next.b].prex=now.a;
                map[next.a][next.b].prey=now.b;
                v[next.a][next.b]=1;
                q.push(next);
            }
        }
        if(now.b!=0)//b->a
        {
            int aa,bb;
            int x=a-now.a;//需要到得量
            if(x>=now.b)
            {
                aa=now.a+now.b;
                bb=0;
                
            }
            else
            {
                aa=a;bb=now.b-x;
            }
            if((!v[aa][bb]||map[aa][bb].step>now.step+1))
            {
                next.a=aa;next.b=bb;
                next.prex=now.a;next.prey=now.b;
                next.step=now.step+1;
                
                map[aa][bb].step=now.step+1;
                strcpy(map[aa][bb].a,"POUR(2,1)");
                map[next.a][next.b].prex=now.a;
                map[next.a][next.b].prey=now.b;
                    v[next.a][next.b]=1;
                q.push(next);
            }
        }
    }
    return false;
}
int main()
{
    while(cin>>a>>b>>c)
    {
        if(!bfs())
            cout<<"impossible"<<endl;
    }
    return 0;
}