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poj2785(4 Values whose Sum is 0)

2016-08-01 17:12 435 查看
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4 Values whose Sum is 0

Time Limit: 15000MS Memory Limit: 228000K
Total Submissions: 19972 Accepted: 5953
Case Time Limit: 5000MS
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 228 ) 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).
Source

Southwestern Europe 2005

题目意思很简单明了，就是从四个数列中各取一个数，使四个数的和为0，求共有多少种方法？

可以直接的四个for循环，但是显然很不高效。把它们分成AB和CD再考虑，就可以快速求解。大致思路是把C、D中的数字枚举出来，排好序，为了使总和为0则需要从C、D中取出c+d=-a-b，可以用二分查找，时间复杂度为O(n^2logn)。

AC代码:

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
#define maxn 4005int A[maxn],B[maxn],C[maxn],D[maxn];
int CD[maxn*maxn];
int main()
{
int n;
scanf("%d",&n);
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++){
CD[i*n+j]=C[i]+D[j];
}
}
sort(CD,CD+n*n);
long long res=0;
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
int cd=-(A[i]+B[j]);
res+=upper_bound(CD,CD+n*n,cd)-lower_bound(CD,CD+n*n,cd);
}
}
printf("%lld\n",res);
return 0;
}

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