POJ -3414-Pots

来源：互联网发布：php相册源码编辑：程序博客网时间：2024/06/14 04:18
题意：
给你，两个容量分别为a,b的杯子，让你看需要几步可以得到c的水，不能的话，impossible，可以的话，输出步数，和路径
思路：
BFS，暴搜
#include<stdio.h>#include<stdlib.h>#include<string.h>#include<iostream>#include<algorithm>#include<queue>using namespace std;int a, b, c;struct node{    int step;    int k;    int n, m;    int qb;} ls[1010],now, next;int vis[101][101];int ki[1010];void bfs(){    memset(vis,0,sizeof(vis));    queue<node> que;    now.step = 0;    now.k = 0;    now.n = 0;    now.m = 0;    que.push(now);    vis[0][0] = 1;    int i;    int ti = 0;    while(!que.empty())    {        now = que.front();        que.pop();        if(now.n == c || now.m == c)        {            printf("%d\n",now.step);            int ko = now.step;            i = 1;            int c = now.qb;            ki[0] = now.k;            while(1)            {                ki[i++] = ls[c].k;                if(i > ko)                    break;                c = ls[c].qb;            }            for(i = ko-1; i >= 0; i--)            {                switch(ki[i])                {                    case 5: printf("FILL(1)\n");break;                    case 6: printf("FILL(2)\n");break;                    case 3: printf("POUR(1,2)\n");break;                    case 4: printf("POUR(2,1)\n");break;                    case 1: printf("DROP(1)\n");break;                    case 2: printf("DROP(2)\n");break;                }            }            return ;        }        ls[++ti] = now;        for(i = 1; i <= 6; i++)        {            if(i == 1)            {                next.n = 0;                next.m = now.m;                next.step = now.step+1;                next.k = i;                if(!vis[next.n][next.m])                {                    vis[next.n][next.m] = 1;                    next.qb = ti;                    que.push(next);                }            }            if(i == 2)            {                next.n = now.n;                next.m = 0;                next.step = now.step+1;                next.k = i;                if(!vis[next.n][next.m])                {                    vis[next.n][next.m] = 1;                    next.qb = ti;                    que.push(next);                }            }            if(i == 3)            {                if(now.n > b-now.m)                {                    next.n = now.n-(b-now.m);                    next.m = b;                }                else                {                    next.n = 0;                    next.m = now.n+now.m;                }                next.k = i;                next.step = now.step+1;                if(!vis[next.n][next.m])                {                    vis[next.n][next.m] = 1;                    next.qb = ti;                    que.push(next);                }            }            if(i == 4)            {                if(now.m > a-now.n)                {                    next.m = now.m-(a-now.n);                    next.n = a;                }                else                {                    next.m = 0;                    next.n = now.m+now.n;                }                next.k = i;                next.step = now.step+1;                if(!vis[next.n][next.m])                {                    vis[next.n][next.m] = 1;                    next.qb = ti;                    que.push(next);                }            }            if(i == 5)            {                next.n = a;                next.m = now.m;                next.step = now.step+1;                next.k = i;                if(!vis[next.n][next.m])                {                    vis[next.n][next.m] = 1;                    next.qb = ti;                    que.push(next);                }            }            if(i == 6)            {                next.n = now.n;                next.m = b;                next.step = now.step+1;                next.k = i;                if(!vis[next.n][next.m])                {                    vis[next.n][next.m] = 1;                    next.qb = ti;                    que.push(next);                }            }        }    }    printf("impossible\n");}int main(){    while(~scanf("%d%d%d",&a,&b,&c))    {        bfs();    }    return 0;}
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