POJ 1840 双向搜索 +Hash

最新推荐文章于 2021-02-26 21:51:37 发布

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#哈希 #搜索 #算法

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本文探讨了一种解决特定形式方程的方法，通过将方程转化为多项式并利用哈希表存储中间结果来优化计算过程，从而高效地计算出满足条件的解的数量。实例演示了如何将方程简化并应用算法进行求解。

Eqs

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 9536		Accepted: 4697

Description

Consider equations having the following form:
a1x1³+ a2x2³+ a3x3³+ a4x4³+ a5x5³=0
The coefficients are given integers from the interval [-50,50].
It is consider a solution a system (x1, x2, x3, x4, x5) that verifies the equation, xi∈[-50,50], xi != 0, any i∈{1,2,3,4,5}.

Determine how many solutions satisfy the given equation.

Input

The only line of input contains the 5 coefficients a1, a2, a3, a4, a5, separated by blanks.

Output

The output will contain on the first line the number of the solutions for the given equation.

Sample Input

37 29 41 43 47

Sample Output

直接搜索复杂度为 100^5 果断坑爹。

转化成 a1x1³+ a2x2³ = a3x3³+ a4x4³+ a5x5³

^{将左边的结果保存到哈希表中。表中的值对应（x1,x2）方案的数目。}

然后枚举右边所有情况。查哈希表。

#include<iostream>
#include<vector>
#include<cstring>
using namespace std;

struct cube{
    cube() {   for(int i=0;i<=50;i++) a[i] = i*i*i; } 
    int operator[](int key)
    {  
        if( key<0 )  return -a[-key];
        else         return a[key]; 
    }
    int a[51];
}Cube;


struct hash{
    enum { MOD = 99997 };
    hash(){ p = 0; memset(h,-1,sizeof(h)); }
    
    struct node{
        int next,key,value;
    }Node[10100];
    int h[MOD],p;
    
    int f(int key)
    { 
        int k=key%MOD;
        return k<0 ? k+MOD : k ;
    }
    int find(int key)
    {
        int k = h[ f(key) ];
        while( k!= -1 ){
            if( Node[k].key == key )  return k;
            k = Node[k].next;
        }
        return -1;
    }
    
    void add(int key)
    {
        int pnode = find(key);
        if( pnode>=0 ){   
            Node[pnode].value++;
        }else{          // not exist
            Node[p].key  = key;
            Node[p].value = 1;
            Node[p].next = h[ f(key) ];
            h[ f(key) ] = p++;
        }
    }
    int value(int key)
    {
        int pnode = find(key);
        if( pnode>=0 ) return Node[pnode].value;
        else           return 0;   // not exist
    }
}Hash;

int main()
{
    int a[5];
    for(int i=0;i<5;i++) cin>>a[i];
   
    for(int i=-50;i<=50;i++){
        if( i==0 ) continue;
        int c0 = Cube[i] * a[0];
        
        for(int j=-50;j<=50;j++){
            if( j==0 ) continue;
            int c1 = c0 + Cube[j]*a[1];
            Hash.add(c1);
        }
    }
    
    int ans = 0;
    for(int i=-50;i<=50;i++){
        if( i==0 ) continue;
        int c2 = Cube[i] * a[2];
        
        for(int j=-50;j<=50;j++){
            if(j==0) continue;
            int c3 = c2 + Cube[j] * a[3];
            
            for(int k=-50;k<=50;k++){
                if( k==0 ) continue;
                int c4 = c3 + Cube[k]*a[4];
                
                ans += Hash.value(c4);
            } // a4
        } // a3
    } // a2
    cout<<ans<<endl;
    
    //system("pause");
    return 0;
}