3414POJ

来源：互联网发布：三菱plc编程教学视频编辑：程序博客网时间：2024/06/07 02:57

给你两个杯子，问你怎么操作可以将这两个杯子其中一个杯子的容量达到目标状态，一共有三个操作

1，把杯子装满。

2，把杯子倒空。

3，一个杯子的水倒向另一杯子。

BFS。。

注意是一次输入一次输入一次输入！！！

我们把每种操作都进行标号，一共有6种，然后每进行一步操作，都记录下来，直到到达目标状态，然后回溯标号就可以。

#include<iostream>#include<cstdio>#include<string.h>#include<string>#include<stack>#include<set>#include<algorithm>#include<cmath>#include<vector>#include<map>#include<sstream>#define ll __int64#define lll unsigned long long#define MAX 1000009#define MAXN 2009#define eps 1e-8#define INF 0x7fffffff#define mod 1000000007#define lson l , m , rt << 1#define rson m + 1 , r , rt << 1 | 1using namespace std;struct node{    int x,y,step,pre,flag;};int v1,v2,c;node queue[MAX];bool vis[109][109];int path[MAX];int index;void bfs(){    memset(vis,false,sizeof(vis));    int front = 0;    int rear = 0;    node u,v;    u.x = 0;    u.y = 0;    u.step = 0;    u.pre = -1;    vis[0][0] = true;    queue[rear++] = u;    while(front!=rear)    {        v = queue[front++];        if(v.x==c||v.y==c)        {            //cout<<v.x<<" "<<v.y<<" "<<v.step<<endl;            break;        }        node next;        for(int i = 0; i<6; i++)        {            if(i==0)            {                next.x = v1;                next.y = v.y;                next.flag = 0;            }            else if(i==1)            {                next.x = v.x;                next.y = v2;                next.flag = 1;            }            else if(i==2)            {                next.x = 0;                next.y = v.y;                next.flag = 2;            }            else if(i==3)            {                next.x = v.x;                next.y = 0;                next.flag = 3;            }            else if(i==4)            {                if(v.x + v.y>=v2)                {                    next.x = v.x - (v2 - v.y);                    next.y = v2;                }                else                {                    next.x = 0;                    next.y = v.x + v.y;                }                next.flag = 4;            }            else if(i==5)            {                if(v.x + v.y>=v1)                {                    next.x = v1;                    next.y = v.y - (v1 - v.x);                }                else                {                    next.x = v.x + v.y;                    next.y = 0;                }                next.flag = 5;            }            if(!vis[next.x][next.y])            {                next.step = v.step + 1;                next.pre = front - 1;                vis[next.x][next.y] = 1;                queue[rear++] = next;            }        }    }    if(front==rear)    {        puts("impossible");        return ;    }    index = 0;    for(int i = front - 1; i>=0;)    {        path[index++] = i;        i = queue[i].pre;    }    printf("%d\n",queue[front-1].step);    for(int i = index - 1; i>=0; i--)    {        int x = queue[path[i]].flag;        switch(x)        {        case 0:            printf("FILL(1)\n");            break;        case 1:            printf("FILL(2)\n");            break;        case 2:            printf("DROP(1)\n");            break;        case 3:            printf("DROP(2)\n");            break;        case 4:            printf("POUR(1,2)\n");            break;        case 5:            printf("POUR(2,1)\n");            break;        }    }}int main(){    cin>>v1>>v2>>c;    bfs();    return 0;}

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