POJ 2785 4 Values whose Sum is 0(折半枚举)

给出四个长度为n的数列a，b，c，d，求从这四个数列中每个选取一个元素后的和为0的方法数。n<=4000,abs(val)<=2^28.

考虑直接暴力，复杂度O(n^4).显然超时。

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-
# define MOD
# define INF
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<,l,mid
# define rch p<<|,mid+,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int res=, flag=;
    char ch;
    if((ch=getchar())=='-') flag=;
    else if(ch>=''&&ch<='') res=ch-'';
    while((ch=getchar())>=''&&ch<='')  res=res*+(ch-'');
    return flag?-res:res;
}
void Out(int a) {
    if(a<) {putchar('-'); a=-a;}
    if(a>=) Out(a/);
    putchar(a%+'');
}
const int N=;
//Code begin...
 
int a[N], b[N], c[N], d[N];
VI vis;
 
int main ()
{
    int n;
    LL ans=;
    n=Scan();
    FOR(i,,n) a[i]=Scan(), b[i]=Scan(), c[i]=Scan(), d[i]=Scan();
    FOR(i,,n) FOR(j,,n) vis.pb(-c[i]-d[j]);
    sort(vis.begin(),vis.end());
    FOR(i,,n) FOR(j,,n) {
        int temp=a[i]+b[j];
        ans+=upper_bound(vis.begin(),vis.end(),temp)-lower_bound(vis.begin(),vis.end(),temp);
    }
    printf("%lld\n",ans);
    return ;
}

枚举a，二分b+c+d.复杂度O(n+n^3*log(n^3)+n*log(n^3))~O(n^3*logn).

枚举a+b,二分b+c.复杂度O(n^2+n^2*log(n^2)+n^2*log(n^2))~O(n^2*logn).

枚举a+b+c,二分d.复杂度O(n^3+logn+n^3*logn)~O(n^3*logn).

另外此题map常数大过不了。