POJ3414解题报告

来源：互联网发布：c语言switch语句用法编辑：程序博客网时间：2024/06/11 09:26
#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;}
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