Poj3414广泛搜索

大家好，又见面了，我是全栈君

<span style="color:#330099;">/*
D - D
Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u
Submit
 
Status
 
Practice
 
POJ 3414
Description
You are given two pots, having the volume of A and B liters respectively. The following operations can be performed:

FILL(i)        fill the pot i (1 ≤ i ≤ 2) from the tap;
DROP(i)      empty the pot i to the drain;
POUR(i,j)    pour from pot i to pot j; after this operation either the pot j is full (and there may be some water left in the pot i), or the pot i is empty (and all its contents have been moved to the pot j).
Write a program to find the shortest possible sequence of these operations that will yield exactly C liters of water in one of the pots.

Input
On the first and only line are the numbers A, B, and C. These are all integers in the range from 1 to 100 and C≤max(A,B).

Output
The first line of the output must contain the length of the sequence of operations K. The following K lines must each describe one operation. If there are several sequences of minimal length, output any one of them. If the desired result can’t be achieved, the first and only line of the file must contain the word ‘impossible’.

Sample Input
3 5 4
Sample Output
6
FILL(2)
POUR(2,1)
DROP(1)
POUR(2,1)
FILL(2)
POUR(2,1)
By Grant Yuan
2014.7.14
poj 3414
广搜
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
using namespace std;
bool flag=0;
int next[6]={0,1,2,3,4,5};
int a,b,c;
int aa,bb,cc;
typedef struct{
  int a;
  int b;
  int f;
  int sum;
  int ope;
}node;
int res;
node q[10000];
bool mark[101][101];
int top,base;
int top1;
int s[10000];
bool can(int x1,int y1)
{
    if(x1>=0&&x1<=aa&&y1>=0&&y1<=bb&&mark[x1][y1]==0)
      return 1;
    return 0;
}

void slove()
{   int a1,b1,f1,a2,b2;
    while(top>=base){//cout<<"zhang"<<endl;
     if(q[base].a==cc||q[base].b==cc){
          flag=1;
           res=q[base].sum;
           break;
          }
      for(int i=0;i<6;i++){
          if(i==0)
            {  a1=aa;
               b1=q[base].b;
               if(can(a1,b1)){
                q[++top].a=a1;
                q[top].b=b1;
                q[top].f=base;
                q[top].sum=q[base].sum+1;
                q[top].ope=i;
                mark[a1][b1]=1;}

            }
            else if(i==1)
            {
               a1=q[base].a;
               b1=bb;
               if(can(a1,b1)){
                q[++top].a=a1;
                q[top].b=b1;
                q[top].f=base;
                q[top].sum=q[base].sum+1;
                q[top].ope=i;
                mark[a1][b1]=1;}
            }
            else if(i==2)//1dao2
            {  int m,n;
               m=q[base].a;
               n=bb-q[base].b;
               if(m>=n){
                   a1=m-n;
                   b1=bb;
                if(can(a1,b1)){
                q[++top].a=a1;
                q[top].b=b1;
                q[top].f=base;
                q[top].sum=q[base].sum+1;
                q[top].ope=i;
                mark[a1][b1]=1;} }
                else{
                  a1=0;
                  b1=m+q[base].b;
                  if(can(a1,b1)){
                q[++top].a=a1;
                q[top].b=b1;
                q[top].f=base;
                q[top].sum=q[base].sum+1;
                q[top].ope=i;
                mark[a1][b1]=1;}
                  }}
            else if(i==3)//1dao2
            {  int m,n;
               m=aa-q[base].a;
               n=q[base].b;
               if(n>=m){
                   a1=aa;
                   b1=n-m;
                if(can(a1,b1)){
                q[++top].a=a1;
                q[top].b=b1;
                q[top].f=base;
                q[top].sum=q[base].sum+1;
                q[top].ope=i;
                mark[a1][b1]=1;} }
                else{
                  b1=0;
                  a1=n+q[base].a;
                  if(can(a1,b1)){
                q[++top].a=a1;
                q[top].b=b1;
                q[top].f=base;
                q[top].sum=q[base].sum+1;
                q[top].ope=i;
                mark[a1][b1]=1;}
                  }}
               else if(i==4)
               {
                   a1=0;
                   b1=q[base].b;
                   if(can(a1,b1)){
                    q[++top].a=a1;
                    q[top].b=b1;
                    q[top].f=base;
                    q[top].sum=q[base].sum+1;
                     q[top].ope=i;
                     mark[a1][b1]=1;
                    }}
                else if(i==5)
               {
                   b1=0;
                   a1=q[base].a;
                   if(can(a1,b1)){
                    q[++top].a=a1;
                    q[top].b=b1;
                    q[top].f=base;
                    q[top].sum=q[base].sum+1;
                    q[top].ope=i;
                     mark[a1][b1]=1;
                    }
               }

        }
        base++;
}}
void print()
{   top1=-1;
    int i=base,j;
    while(1){
        s[++top1]=q[i].ope;
        j=q[i].f;
        i=j;
        if(i==0)
          break;
          }
    for(j=top1;j>=0;j--)
    {
        if(s[j]==0)
           cout<<"FILL(1)"<<endl;
        else
          if(s[j]==1)
           cout<<"FILL(2)"<<endl;
        else
           if(s[j]==2)
            cout<<"POUR(1,2)"<<endl;
        else
          if(s[j]==3)
            cout<<"POUR(2,1)"<<endl;
        else
          if(s[j]==4)
            cout<<"DROP(1)"<<endl;
        else
          if(s[j]==5)
            cout<<"DROP(2)"<<endl;
    }
}
int main()
{
    cin>>aa>>bb>>cc;
    top=-1;
    memset(mark,0,sizeof(mark));
    mark[0][0]=1;
    base=0;
    q[++top].a=0;
    q[top].b=0;
    q[top].f=0;
    q[top].sum=0;
    q[top].ope=0;
    slove();
    if(flag==0) cout<<"impossible"<<endl;
    else{cout<<res<<endl;
    print();}
    return 0;

}
</span>

复制

版权声明：本文博客原创文章。博客，未经同意，不得转载。

发布者：全栈程序员栈长，转载请注明出处：https://javaforall.cn/117579.html原文链接：https://javaforall.cn