（HDU - 4334）Trouble

（HDU - 4334）Trouble

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 6218 Accepted Submission(s): 1738

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

题目大意：在5个集合中分别取一个数使这5个数的和为0。

思路：这题刚开始我想到的是二分，因为这题就是典型的二分，先预处理出前三个集合的和，后两个集合的和，然后二分查找，时间复杂度为O(n3logn)。再看了大佬的思路后，发现了一种更好的方法，先利用分治思想，处理前两个集合的和sum1，3 4集合的和sum2。然后要介绍一种尺取法(也就是一种贪心思想)：如何快速判断是否有A[i]+B[j]=x，先将A,B送小到大排序得到有序序列，可以先让i指向A数组的首指针，j指向B数组的为指针，判断A[i]+B[j]与x的大小关系，若A[i]+B[j]>x则j–；若A[i]+B[j]< x则i++。这样就可以在线性时间内判断是否有A[i]+B[j]=x。利用这种方法最后的复杂度为O(n3)。显然选择这种方法更好。

#include<cstdio>
#include<algorithm>
using namespace std;

typedef long long LL;
const int maxn=205;
LL a[5][maxn];
LL sum1[maxn*maxn],sum2[maxn*maxn];

int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n;
scanf("%d",&n);
for(int i=0;i<5;i++)
for(int j=0;j<n;j++) scanf("%lld",&a[i][j]);
int k=0;
for(int i=0;i<n;i++)
for(int j=0;j<n;j++) sum1[k++]=a[0][i]+a[1][j];
k=0;
for(int i=0;i<n;i++)
for(int j=0;j<n;j++) sum2[k++]=a[2][i]+a[3][j];
sort(sum1,sum1+k);
sort(sum2,sum2+k);
bool flag=false;
for(int i=0;i<n&&!flag;i++)
{
int l=0,r=k-1;
while(l<k&&r>=0)
{
if(sum1[l]+sum2[r]==-a[4][i])
{
flag=true;
break;
}
else if(sum1[l]+sum2[r]>-a[4][i]) r--;
else l++;
}
}
if(flag) printf("Yes\n");
else printf("No\n");
}
return 0;
}