宽度优先搜索模拟倒水

B - Pots

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

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

6FILL(2)POUR(2,1)DROP(1)POUR(2,1)FILL(2)POUR(2,1)

 

想法：

    这个题明显样本容量并不很大而且算法本人也忘掉了，所以可以使用暴力搜索或者打表（反正不大）

   下面转载的是来自一位博主的文章...（初学者):

这里是利用数组进行的宽搜

#include<stdio.h>#include<string.h>const int maxn = 110;int vis[maxn][maxn]; //标记状态是否入队过int a,b,c; //容器大小int step; //最终的步数int flag; //纪录是否能够成功/* 状态纪录 */struct Status{    int k1,k2; //当前水的状态    int op; //当前操作    int step; //纪录步数    int pre; //纪录前一步的下标}q[maxn*maxn];int id[maxn*maxn]; //纪录最终操作在队列中的编号int lastIndex; //最后一个的编号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(){    while(scanf("%d%d%d", &a,&b,&c) != EOF)    {        flag = 0; //初始化不能成功        step = 0;        bfs();        if(flag)        {            printf("%d\n", step);            id[step] = lastIndex; //最后一步在模拟数组中的编号            for(int i = step-1; i >= 1; i--)            {                id[i] = q[id[i+1]].pre; //向前找前一步骤在模拟数组中的编号            }            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;}