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

221 -11 -11 -11 -11 -131 2 3-1 -2 -34 5 6-1 3 2-4 -10 -1

 

Sample Output

NoYes

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;}