POJ_2758 (二分)

 

Description

 

The SUM problem can be formulated as follows: given four lists A, B, C, D of integer values, compute how many quadruplet (a, b, c, d ) ∈ A x B x C x D are such that a + b + c + d = 0 . In the following, we assume that all lists have the same size n .

 

Input

The first line of the input file contains the size of the lists n (this value can be as large as 4000). We then have n lines containing four integer values (with absolute value as large as 2 ²⁸ ) that belong respectively to A, B, C and D .

 

Output

For each input file, your program has to write the number quadruplets whose sum is zero.

 

Sample Input

6
-45 22 42 -16
-41 -27 56 30
-36 53 -37 77
-36 30 -75 -46
26 -38 -10 62
-32 -54 -6 45

Sample Output

5

Hint

Sample Explanation: Indeed, the sum of the five following quadruplets is zero: (-45, -27, 42, 30), (26, 30, -10, -46), (-32, 22, 56, -46),(-32, 30, -75, 77), (-32, -54, 56, 30).

 

题意为从给出的四列数中，每列选出一个数，可以重复选，问有多少组这四个数相加结果等于0。

每列的个数最多四千 ，暴力4000^4，一定超时的。我们中的巨巨想到一个办法。先将数组预处理下两个两个合并，前两个合并，后两个合并。这样新生成的两个数组，第一个为前两列各个数字的和，第二个为后两列各个数字的和。这样的复杂度为O（n^2）,16000000,这样不会超时，接下来遍历其中一个数组，二分另外一个数组，如果遍历的那个数组元素的相反数等于，在另外的一个数组中二分找到了，ans++；

这个思路是很厉害的，可是交上去一直wa，我们各种改还是不对，比赛结束后才想到，万一二分的那个数组中要查找的数出现了多次怎么办！漏洞就在这，把查找到的那个数在向两边搜下看有没有相等的就好了，注意break；我在这超时了两发~~~~~

 

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 4005;
const int maxnn = 16000000+25;
int a[maxn],b[maxn],c[maxn],d[maxn],ab[maxnn],cd[maxnn];
int n,m,k;

int s_division(int x){
	int l,r,mid,ans;
	l = 0; r = m-1; mid = ans = 0;
	while(l<=r){
		mid=(l+r)>>1;
		if(cd[mid]==x){
			ans++;
			for(int j=mid+1;j<m;j++){
				if(cd[mid]==cd[j])
					ans++;
                                else break;
			}
			for(int j=mid-1;j>=0;j--){
				if(cd[mid]==cd[j])
					ans++;
                                else break;
			}
			return ans;
		}
		if(cd[mid]<x)
			l=mid+1;
		if(cd[mid]>x)
			r=mid-1;
	}
	return ans;	
}

int main(){
	int n,sum;
	while(scanf("%d",&n)==1){
		sum = 0;k = m = 0;
		for(int i=0;i<n;i++){
			scanf("%d %d %d %d",&a[i],&b[i],&c[i],&d[i]);
		}
		for(int i=0;i<n;i++){
			for(int j=0;j<n;j++){
				ab[k++]=a[i]+b[j];
			}
		}
		for(int i=0;i<n;i++){
			for(int j=0;j<n;j++){
				cd[m++]=c[i]+d[j];
			}
		}
		sort(cd,cd+m);
		for(int i=0;i<k;i++){
			sum+=s_division(-1*ab[i]);
		}
		printf("%d\n",sum);
	}
	return 0;
}