HDU 4334 Trouble (数组合并)

Trouble
Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5681    Accepted Submission(s):
1570


[align=left]Problem Description[/align]
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?
 
 

[align=left]Input[/align]
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.
 
 

[align=left]Output[/align]
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".
 
 

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

No
Yes

题意：t组数据，每组输入一个n，接下来有五行代表五个集合，每行n个数，问是否能从每个集合中找出一个数，
　　　使得和为0。
题解：直接暴力会超时，将12两个集合合并，34两个集合合并，然后对他们进行从小到大排序，跑第五个集合。
　　　用两个指针下标将12从小到大遍历，34逆序从大到小遍历，如果和<0，12集合的指针后移，如果和>0，
　　　34集合的指针左移，如果和为0标记flag=1;break一下输出就行了。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
long long a[40005],b[40005],c[205],d[205],e[205],f[205],g[205];
long long i,j,x,k,l,n,t,flag=0,p;
int main()
{
scanf("%lld",&t);
while(t--)
{
flag=0;
//输入
scanf("%lld",&n);
for(i=0;i<n;i++)
scanf("%lld",&d[i]);
for(i=0;i<n;i++)
scanf("%lld",&e[i]);
for(i=0;i<n;i++)
scanf("%lld",&f[i]);
for(i=0;i<n;i++)
scanf("%lld",&g[i]);
for(i=0;i<n;i++)
scanf("%lld",&c[i]);
//合并
int cnt1=0;
for(i=0;i<n;i++)
for(j=0;j<n;j++)
a[cnt1++]=d[i]+e[j];
int cnt2=0;
for(i=0;i<n;i++)
for(j=0;j<n;j++)
b[cnt2++]=f[i]+g[j];  //三四合并
sort(a,a+cnt1);
sort(b,b+cnt2);
//计算
for(i=0;i<n;i++) //第五个数组 c[]
{
j=0;
k=cnt2-1;
while(j<cnt1 && k>=0)
{
if(a[j]+b[k]+c[i]==0LL)
{
flag=1;
break;
}
else if(a[j]+b[k]+c[i]<0)  //注意不要都写成i
j++;
else
k--;
}

}
if(flag) cout<<"Yes"<<endl; //注意是Yes不是YES也不是yes
else cout<<"No"<<endl;
}
return 0;
}