hdu 4334 Trouble

Problem Description

Hassan is in trouble. His mathematics teacher has given him a very difficult problem called 5-sum. Please help him.
The 5-sum problem is defined as follows: Given 5 sets S_1,...,S_5 of n integer numbers each, is there a_1 in S_1,...,a_5 in S_5 such that a_1+...+a_5=0?

 

Input

First line of input contains a single integer N (1≤N≤50). N test-cases follow. First line of each test-case contains a single integer n (1<=n<=200). 5 lines follow each containing n integer numbers in range [-10^15, 1 0^15]. I-th line denotes set S_i for 1<=i<=5.

 

Output

For each test-case output "Yes" (without quotes) if there are a_1 in S_1,...,a_5 in S_5 such that a_1+...+a_5=0, otherwise output "No".

 

Sample Input

  
  

   
   2
2
1 -1
1 -1
1 -1
1 -1
1 -1
3
1 2 3
-1 -2 -3
4 5 6
-1 3 2
-4 -10 -1
  
  

 

Sample Output

  
  

   
   No
Yes
  
  

思想很暴力，纯暴力的方法为n的5次方，显然不能过。用hash进行离散优化（比赛的时候写的二分果断TLE）

hash的复杂度是n的3次方，而二分的复杂度是n的三次方乘以15左右。

将第一行加上第二行的值的相反数起来存入hash表中，然后查询第三行加第四行加第五行的值是否在 hash表中出现，如果出现则为yes，否则为no

#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<vector>
using namespace std;
const int maxn=210;
const int MOD=1000007;//hash离散化，模上一个数字 
struct in
{
    __int64 val;//保存当前的值 
    int next;//因为存在负数，也就是同一个tmp保存了两个值 
}hash[maxn*maxn];

__int64 org[6][maxn];
int head[MOD];
int N,n;
int tot;
void init()
{
    tot=0;
    memset(head,-1,sizeof(head));
}
void add_hash(__int64 val)
{
    int tmp=val%MOD;
    tmp=tmp>=0?tmp:-tmp;
    for(int i=head[tmp];i!=-1;i=hash[i].next)//i判定读入的值之前是否有读入，head[tmp]保存出现值的位置 
                                             // i最多为两个值，tmp为正数时的位置和 tmp为负数时的位置 
    {
        if(hash[i].val==val)
        return;
    }
    hash[tot].val=val;
    hash[tot].next=head[tmp];
    head[tmp]=tot++;//有新值读入才++ 
    
}
bool query(__int64 val)
{
    int tmp=val%MOD;
    tmp=tmp>=0?tmp:-tmp;
    for(int i=head[tmp];i!=-1;i=hash[i].next)
    {
        if(hash[i].val==val)
        return true;
    }
    return false;
}
bool slove()
{
    int i,j,k;
    for( i=0;i<n;i++)
    {
        for( j=0;j<n;j++)
        {
            for( k=0;k<n;k++)
            {
                if(query((org[2][i]+org[3][j]+org[4][k])))
                return true;
            }
        }
    }
    return false;
}
int main()
{
    scanf("%d",&N);
        while(N--)
        {
            scanf("%d",&n);
            init();
            for(int i=0;i<5;i++)
            {
                for(int j=0;j<n;j++)
                scanf("%I64d",&org[i][j]);
            }
            for(int i=0;i<n;i++)
            {
                for(int j=0;j<n;j++)
                add_hash(-(org[0][i]+org[1][j]));
            }
            if(slove())
            {
                puts("Yes");
            }
            else 
            {
                puts("No");
            }
        }
        
        
    
    return 0;
}