折半枚举（双向搜索）poj27854 Values whose Sum is 0
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4 Values whose Sum is 0
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).
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有时候问题的规模比较大，无法枚举所有元素的组合，但能够枚举一般元素的组合。此时，将问题拆成两半后分别枚举，再合并他们的结果这一方法往往非常有效。
//折半枚举（双向搜索）poj2785
#include <iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn=5005;
int n;
ll a[maxn],b[maxn],c[maxn],d[maxn];
ll cd[maxn*maxn];
void solve()
{
//枚举cd的组合
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);
ll res=0;
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
ll CD=-(a[i]+b[j]);
//二分搜索取出cd中和为CD的部分
res+=upper_bound(cd,cd+n*n,CD)-lower_bound(cd,cd+n*n,CD);
}
}
printf("%lld\n",res);
}
int main()
{
cin>>n;
for(int j=0;j<n;j++)
{
cin>>a[j]>>b[j]>>c[j]>>d[j];
}
solve();
return 0;
}
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