3414POJ

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 | 1

using 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;
}