poj 3414 Pots【bfs+回溯路径 正向输出】

题目地址：http://poj.org/problem?id=3414

Pots

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)

#include<stdio.h>
#include<string.h>
#include <queue>
#include <stack>
#include <algorithm>

using namespace std;

int a, b, c;//a b容器的大小 c目标体积

bool vis[101][101];

struct node
{
    char ch;
    int x,y;
    int pre;
    int self;
}op[100000];
int e=0;

struct path
{
    int u, v;
    int cnt;
    int pre;
    int self;
};

//int road[10000], step;

stack<int>p;
bool ok;
int ans;

void bfs()
{
    queue<path>q;//创建队列
    memset(vis, false, sizeof(vis));//初始化未访问

    path s; s.u=0; s.v=0; s.cnt=0; s.pre=-1; s.self=-1;
    q.push(s);  int num=-1;
    vis[0][0]=true; ok=false; ans=0;

    while(!q.empty())
    {
        path cur=q.front(); q.pop();

        if(cur.u==c || cur.v==c){
            ans=cur.cnt;
            ok=true;
            int fa=cur.self;
            while(fa!=-1){
                p.push(fa);
                fa=op[fa].pre;
            }
            break;
        }//如果找到了目标状态

        //进行6种操作
        path temp;
        //Fill A
        temp=cur; temp.u=a; temp.cnt=cur.cnt+1;
        if(!vis[temp.u][temp.v]){
            temp.pre=cur.self; num++; temp.self=num;
            q.push(temp);
            vis[temp.u][temp.v]=true;//标记状态访问
            op[e].pre=cur.self;
            op[e].ch=‘F‘; op[e++].x=1;
        }

        //File B
        temp=cur; temp.v=b; temp.cnt=cur.cnt+1;
        if(!vis[temp.u][temp.v]){
            temp.pre=cur.self; num++; temp.self=num;
            q.push(temp);
            vis[temp.u][temp.v]=true;
            op[e].pre=cur.self;
            op[e].ch=‘F‘; op[e++].x=2;
        }

        //Drop A
        temp=cur; temp.u=0; temp.cnt=cur.cnt+1;
        if(!vis[temp.u][temp.v]){
            temp.pre=cur.self; num++; temp.self=num;
            q.push(temp);
            vis[temp.u][temp.v]=true;
            op[e].pre=cur.self;
            op[e].ch=‘D‘; op[e++].x=1;
        }
        //Drop B
        temp=cur; temp.v=0; temp.cnt=cur.cnt+1;
        if(!vis[temp.u][temp.v]){
            temp.pre=cur.self; num++; temp.self=num;
            q.push(temp);
            vis[temp.u][temp.v]=true;
            op[e].pre=cur.self;
            op[e].ch=‘D‘; op[e++].x=2;
        }

        //pour(A, B)
        temp.v=cur.u+cur.v; temp.u=0; temp.cnt=cur.cnt+1;
        if(temp.v>b){
            temp.v=b;//装满B
            temp.u=cur.u+cur.v-b;
        }
        if(!vis[temp.u][temp.v]){
            temp.pre=cur.self; num++; temp.self=num;
            q.push(temp);
            vis[temp.u][temp.v]=true;
            op[e].pre=cur.self;
            op[e].ch=‘P‘; op[e].x=1; op[e++].y=2;
        }

        //pour(B, A)
        temp.u=cur.u+cur.v; temp.v=0; temp.cnt=cur.cnt+1;
        if(temp.u>a){
            temp.u=a;//装满A
            temp.v=cur.u+cur.v-a;
        }
        if(!vis[temp.u][temp.v]){
            temp.pre=cur.self; num++; temp.self=num;
            q.push(temp);
            vis[temp.u][temp.v]=true;
            op[e].pre=cur.self;
            op[e].ch=‘P‘; op[e].x=2; op[e++].y=1;
        }

    }
}

int main()
{
    scanf("%d %d %d", &a, &b, &c);
    e=0;
    bfs();

    if(!ok){
        printf("impossible\n");
    }
    else{
        printf("%d\n", ans);
        while(!p.empty())
        {
            int dd=p.top(); p.pop();

            if(op[dd].ch==‘F‘){
                printf("FILL(%d)\n", op[dd].x );
            }
            else if(op[dd].ch==‘D‘){
                printf("DROP(%d)\n", op[dd].x );
            }
            else{
                printf("POUR(%d,%d)\n", op[dd].x, op[dd].y );
            }
        }
    }

	return 0;
}

POUR(2,1) DROP(1) POUR(2,1) FILL(2) POUR(2,1)

代码：