HDU 4334 Trouble

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Trouble

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5678    Accepted Submission(s): 1568

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
   
   
   
   

 

Source

2012 Multi-University Training Contest 4

 

题目大意：

就是给定5个集合 让你判断是否这5个集合中的数 能够相加==0

解题思路：

就是两个两个集合 合并 排序，然后剩下一个集合跑， 还得去重（不去也行，去重循环少点），将重新组合后的3个集合，让没有合并的集合放在最外层循环，然后里边就是一个从前往后跑，一个从后往前跑，如果相加>0,那个降序排列的-1，如果<0,升序的加+1。

PS：其实我以为暴力是不行的 会超时，应该是数据不是很强吧 哈哈~ 最坏的情况是 200*200*200*50 = 4 * 10^8，反正这样就过了。。。还是得看看正解呀

My Code：

include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long LL;
const int MAXN = 4e4+5;
LL a[205],b[205],c[205],d[205],e[205];
LL t1[MAXN],t2[MAXN];
int main()
{
    int n, T;
    cin>>T;
    while(T--)
    {
        scanf("%d",&n);
        for(int i=0; i<n; i++)
            scanf("%lld",&a[i]);
        for(int i=0; i<n; i++)
            scanf("%lld",&b[i]);
        for(int i=0; i<n; i++)
            scanf("%lld",&c[i]);
        for(int i=0; i<n; i++)
            scanf("%lld",&d[i]);
        for(int i=0; i<n; i++)
            scanf("%lld",&e[i]);
        int cnt1=0, cnt2=0;
        for(int i=0; i<n; i++)
            for(int j=0; j<n; j++)
                t1[cnt1++] = a[i] + b[j];
        for(int i=0; i<n; i++)
            for(int j=0; j<n; j++)
                t2[cnt2++] = c[i] + d[j];
        sort(t1,t1+cnt1);
        sort(t2,t2+cnt2);
        cnt1 = unique(t1,t1+cnt1)-t1;
        cnt2 = unique(t2,t2+cnt2)-t2;
        bool ok = 0;
        for(int i=0; i<n; i++)
        {
            for(int j=0,k=cnt2-1; j<cnt1&&k>=0;)
            {
                if(t1[j]+t2[k]+e[i] == 0)
                {
                    ok = 1;
                    break;
                }
                else if(t1[j]+t2[k]+e[i] > 0)
                    k--;
                else
                    j++;
            }
            if(ok)
            {
                puts("Yes");
                break;
            }
        }
        if(!ok)
            puts("No");
    }
    return 0;
}