H, cartoon h

H, cartoon h

This question is not very difficult, but due to some details, I spent two whole hours debugging;

Later, xuejie helped me find the problem. It took a long time to find out the problem. Later, I must pay attention to the details.

///////////////////////////////////

The idea is that there are a total of two jars, each of which can be two operations, a total of six operations, on which a wide search can be performed.

#include <stdio.h>#include <algorithm>#include <string.h>#include <math.h>#include <queue>#include <vector>#include <iostream>using namespace std;#define N 2000char a[10][100] ={ "FILL(1)","FILL(2)","DROP(1)","DROP(2)","POUR(1,2)","POUR(2,1)"};struct node{    int x, y, t;    char op[N];};int v[N][N];int n, m, k;void bfs(){    node q, p;    queue<node>Q;    p.x=0;    p.y=0;    p.t=0;    p.op[0]='\0';    v[0][0]=1;    Q.push(p);    while(Q.size())    {        p=Q.front();        Q.pop();        if(p.x==k || p.y==k)        {            printf("%d\n", p.t);            for(int i=0;i<p.t;i++)                printf("%s\n",a[p.op[i]-'0']);            return;        }        for(int i=0;i<6;i++)        {            q.t = p.t;            q.x = p.x;q.y = p.y;            strcpy(q.op, p.op);            if(i==0)            {                q.x=n,q.y = p.y;                q.op[q.t] = '0';            }            else if(i==1)            {                q.y=m,q.x = p.x;                q.op[q.t] = '1';            }            else if(i==2)            {                q.x=0,q.y = p.y;                q.op[q.t] = '2';            }            else if(i==3)            {                q.y=0,q.x = p.x;                q.op[q.t] = '3';            }            else if(i==4)            {                if(q.x+q.y<=m)                {                    q.y+=q.x;                    q.x=0;                }                else                {                    q.x=q.x+q.y-m;                    q.y=m;                }                q.op[q.t] = '4';            }            else if(i==5)            {                if(q.x+q.y<=n)                {                    q.x+=q.y;                    q.y=0;                }                else                {                    q.y=q.x+q.y-n;                    q.x=n;                }                q.op[q.t] = '5';            }            if(v[q.x][q.y]==0)            {                v[q.x][q.y]=1;                q.t++;                q.op[q.t]='\0';                Q.push(q);            }        }    }    printf("impossible\n");}int main(){    while(scanf("%d%d%d", &n, &m, &k)!=EOF)    {        memset(v, 0, sizeof(v));        bfs();    }    return 0;}

 

 

Sample Input

3 5 4

Sample Output

6FILL(2)POUR(2,1)DROP(1)POUR(2,1)FILL(2)POUR(2,1)