HDU 4334 Trouble

Trouble

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3526    Accepted Submission(s): 1113

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

 

题意：

给定五个集合，从五个集合中分别取出5个数，如果存在满足a1 +a2 + a3 + a4 +a5 = 0

分析：

事实上考虑如下问题，快速求解序列A，序列B中，是否有Ai+Bj=x ,记得是某知名公司的一道面试题吧；

是两个指针的应用，将A，B升序排列，初试 i 指针指向A[1] ，j 指针指向 B[b_size] ,比较Ai + Bj 与 x 的

大小。若A[i]+B[j]<x , i 指针右移 ; 若 A[i]+B[j]>x , j 指针左移 。事实上是一个贪心的思想。

个人感悟：

使用returen 比 break 好吧。

#include <iostream>
#include <string.h>
#include <string>
#include <algorithm>
#include <stdio.h>
#include <queue>
#include <set>
#define Max(a,b) ((a)>(b)?(a):(b))
using namespace std ;
typedef long long LL ;
struct Me{
   LL N ,N_2 ,a_size ,b_size ,c_size;
   LL num[][] ;
   LL A[*] ;
   LL B[*] ;
   LL C[] ;
   Me(){}
   Me(int n):N(n){}
   void read_init(){
     for(int i=;i<=;i++)
       for(int j=;j<=N;j++)
          scanf("%I64d",&num[i][j]) ;
     int k ;
     k=;
     for(int i=;i<=N;i++)
        for(int j=;j<=N;j++)
           A[++k]=num[][i]+num[][j];
     k= ;
     for(int i=;i<=N;i++)
        for(int j=;j<=N;j++)
           B[++k]=num[][i]+num[][j];
     for(int i=;i<=N;i++)
         C[i]=num[][i] ;
   }
   int my_search(LL x){
       int i= ;
       int j=b_size ;
       while(i<=a_size&&j>=){
          if(A[i]+B[j]<x)
            i++ ;
          else if(A[i]+B[j]==x)
            return  ;
          else if(A[i]+B[j]>x)
            j-- ;
       }
       return  ;
   }
   int calc(){
      sort(A+,A++N*N) ;
      a_size=unique(A+,A++N*N)-(A+) ;
      sort(B+,B++N*N) ;
      b_size=unique(B+,B++N*N)-(B+) ;
      sort(C+,C++N) ;
      c_size=unique(C+,C++N)-(C+) ;
      for(int i=;i<=c_size;i++){
            if(my_search(-*C[i]))
                return  ;
      }
      return  ;
   }
   void gao_qi(){
      read_init() ;
      if(calc())
        puts("Yes") ;
      else
        puts("No") ;
   }
};
int main(){
  int T ,N ;
  cin>>T ;
  while(T--){
    scanf("%d",&N) ;
    Me me(N) ;
    me.gao_qi() ;
  }
  return  ;
}