POJ3414解题报告

#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[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;
}