POJ3414 Pots

题目链接：https://vjudge.net/problem/POJ-3414
Description
You are given two pots, having the volume of A and B liters respectively. The following operations can be performed:
FILL(i) fill the pot i (1 ≤ i ≤ 2) from the tap;
DROP(i) empty the pot i to the drain;
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)

题目大意：给你两个容器，分别能装下A升水和B升水，并且可以进行以下操作
FILL(i) 将第i个容器从水龙头里装满(1 ≤ i ≤ 2);
DROP(i) 将第i个容器抽干
POUR(i,j) 将第i个容器里的水倒入第j个容器（这次操作结束后产生两种结果，一是第j个容器倒满并且第i个容器依旧有剩余，二是第i个容器里的水全部倒入j中，第i个容器为空）
现在要求你写一个程序，来找出能使其中任何一个容器里的水恰好有C升，找出最少操作数并给出操作过程

Cpp Code

#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
const int maxn = 110;
bool vis[maxn][maxn];//标记是否入队过
int a, b, c, step, flag;
struct Status{
    int k1, k2, op, step, pre;
    /*k1 k2当前水的状态
      op 当前操作
      step 步数   
      pre 前一个状态的下标*/
} q[maxn * maxn];
int id[maxn * maxn];//记录最终操作在队列中的编号
int lastIndex;

void bfs(){
    Status now, next;

    int head, tail;
    head = tail = 0;

    q[tail].k1 = 0; q[tail].k2 = 0;
    q[tail].op = 0; q[tail].step = 0; q[tail].pre = 0;

    tail++;

    memset(vis,0,sizeof(vis));
    vis[0][0] = 1; //标记初始状态已入队

    while(head < tail) //当队列非空
    {
        now = q[head]; //取出队首
        head++; //弹出队首

        if(now.k1 == c || now.k2 == c) //应该不会存在这样的情况, c=0
        {
            flag = 1;
            step = now.step;
            lastIndex = head-1; //纪录最后一步的编号
        }

        for(int i = 1; i <= 6; i++) //分别遍历 6 种情况
        {
            if(i == 1) //fill(1)
            {
                next.k1 = a;
                next.k2 = now.k2;
            }
            else if(i == 2) //fill(2)
            {
                next.k1 = now.k1;
                next.k2 = b;
            }
            else if(i == 3) //drop(1)
            {
                next.k1 = 0;
                next.k2 = now.k2;
            }
            else if(i == 4) // drop(2);
            {
                next.k1 = now.k1;
                next.k2 = 0;
            }
            else if(i == 5) //pour(1,2)
            {
                if(now.k1+now.k2 <= b) //如果不能够装满 b
                {
                    next.k1 = 0;
                    next.k2 = now.k1+now.k2;
                }
                else //如果能够装满 b
                {
                    next.k1 = now.k1+now.k2-b;
                    next.k2 = b;
                }
            }
            else if(i == 6) // pour(2,1)
            {
                if(now.k1+now.k2 <= a) //如果不能够装满 a
                {
                    next.k1 = now.k1+now.k2;
                    next.k2 = 0;
                }
                else //如果能够装满 b
                {
                    next.k1 = a;
                    next.k2 = now.k1+now.k2-a;
                }
            }

            next.op = i; //纪录操作
            if(!vis[next.k1][next.k2]) //如果当前状态没有入队过
            {
                vis[next.k1][next.k2] = 1; //标记当前状态入队
                next.step = now.step+1; //步数 +1
                next.pre = head-1; //纪录前一步的编号

                //q.push(next);
                //q[tail] = next; 加入队尾
                q[tail].k1 = next.k1; q[tail].k2 = next.k2;
                q[tail].op = next.op; q[tail].step = next.step; q[tail].pre = next.pre;
                tail++; //队尾延长

                if(next.k1 == c || next.k2 == c) //如果达到目标状态
                {
                    flag = 1; //标记成功
                    step = next.step; //纪录总步骤数
                    lastIndex = tail-1; //纪录最后一步在模拟数组中的编号
                    return;
                }
            }
        }
    }
}
int main(){
    cin >> a >> b >> c;
    flag = 0;
    step = 0;
    bfs();
    if(flag){
        printf("%d\n", step);
        id[step] = lastIndex;
        for (int i = step - 1; i >= 1;i--){
            id[i] = q[id[i + 1]].pre;
        }
        for (int i = 1; i <= step;i++){
            if(q[id[i]].op == 1)
                    printf("FILL(1)\n");

                else if(q[id[i]].op == 2)
                    printf("FILL(2)\n");

                else if(q[id[i]].op == 3)
                    printf("DROP(1)\n");

                else if(q[id[i]].op == 4)
                    printf("DROP(2)\n");

                else if(q[id[i]].op == 5)
                    printf("POUR(1,2)\n");

                else if(q[id[i]].op == 6)
                    printf("POUR(2,1)\n"); 
        }
    }
    else{
        printf("impossible\n");
    }
    return 0;
}