poj 3414 路径输出的bfs（pre)

#include <iostream>  

#include <algorithm>  

using namespace std;  

#include <cstring>  

#include <queue>  

#include <stack>  

struct cup  

{  

    int x, y;  

    int step;  

    int flag;//标记操作  

    cup *pre;//记录路径  

};  

queue<cup>Q;  

stack<int>R;  

int a, b, e;  

int vis[117][117]={0};//标记当前状态是否到达过  

int ans;  

void BFS(int x, int y)  

{  

    cup c;  

    cup t[317];//目前瓶子里剩余的水量  

    c.x = 0, c.y = 0;  

    c.flag = 0;  

    c.pre = NULL;  

    c.step = 0;  

    Q.push(c);  

    vis[x][y] = 1;  

    int count = -1;  

    while(!Q.empty())  

    {  

        count++;  

        t[count] = Q.front();  

        Q.pop();  

        for(int i = 1; i <= 6; i++)  

        {  

            switch(i)  

            {  

                case 1:                     //fill a  

                    c.x = a;  

                    c.y = t[count].y;  

                    c.flag = 1;  

                    break;  

                case 2:                     //fill b  

                    c.x = t[count].x;  

                    c.y = b;  

                    c.flag = 2;  

                    break;  

                case 3:                     //drop a  

                    c.x = 0;  

                    c.y = t[count].y;  

                    c.flag = 3;  

                    break;  

                case 4:                     //drop b  

                    c.x = t[count].x;  

                    c.y = 0;  

                    c.flag = 4;  

                    break;  

                case 5:                     //pour a to b  

                    if(t[count].x > b-t[count].y)  

                    {  

                        c.x = t[count].x-(b-t[count].y);  

                        c.y = b;  

                    }  

                    else  

                    {  

                        c.x = 0;  

                        c.y = t[count].y+t[count].x;  

                    }  

                    c.flag = 5;  

                    break;  

                case 6:                     //pour b to a  

                    if(t[count].y > a-t[count].x)  

                    {  

                        c.y = t[count].y - (a-t[count].x);  

                        c.x = a;  

                    }  

                    else  

                    {  

                        c.x = t[count].x+t[count].y;  

                        c.y = 0;  

                    }  

                    c.flag = 6;  

                    break;  

            }  

            if(vis[c.x][c.y])  

                continue;  

            vis[c.x][c.y] = 1;  

            c.step = t[count].step+1;  

            c.pre = &t[count];  

            if(c.x == e || c.y == e)  

            {  

                ans = c.step;  

                while(c.pre)  

                {  

                    R.push(c.flag);  

                    c = *c.pre;  

                }  

                return;  

            }  

            Q.push(c);  

        }  

    }  

}  

void print()  

{  

    while(!R.empty())  

    {  

        int i = R.top();  

        R.pop();  

        switch(i)  

        {  

            case 1:cout<<"FILL(1)"<<endl;break;  

            case 2:cout<<"FILL(2)"<<endl;break;  

            case 3:cout<<"DROP(1)"<<endl;break;  

            case 4:cout<<"DROP(2)"<<endl;break;  

            case 5:cout<<"POUR(1,2)"<<endl;break;  

            case 6:cout<<"POUR(2,1)"<<endl;break;  

        }  

    }  

}  

int main()  

{  

    cin >>a>>b>>e;  

    BFS(0,0);  

    if(ans == 0)  

        cout<<"impossible"<<endl;  

    else  

    {  

        cout<<ans<<endl;  

        print();  

    }  

    return 0;  

}