Semi-enumeration (bidirectional search)

There are four Series A, B, C, and D with N integers. Take a number from each sequence so that the sum of the four numbers is 0. Find the number of such combinations.

Input

N = 6

A = {-45,-41,-36,-36, 26,-32}

B = {22,-27, 53, 30,-38,-54}

C = {42, 56,-37,-75,-10,-6}

D = {-16, 30, 77,-46, 62, 45}

Select a total of N4 cases from the four columns, and all cases are judged as unfeasible. However, if we divide them into AB and CD, we can solve the problem quickly. If you select only N2 combinations from two columns, You can enumerate them. After extracting a and B From A and B, C + D =-(A + B) needs to be extracted from C and D to make the sum 0 ). Therefore, the N2 methods that retrieve numbers from C and D are enumerated and sorted to perform binary search. The complexity of this algorithm is O (n2logn ).

1 # include <iostream> 2 # include <algorithm> 3 # include <cstdio> 4 using namespace STD; 5 6 # define max_n 1000 7 8 int N; 9 int A [max_n], B [max_n], C [max_n], d [max_n]; 10 int CD [max_n * max_n]; // combination of numbers in C and D 11 12 Void solve () 13 {14 // enumeration all methods for extracting numbers from C and D 15 for (INT I = 0; I <n; I ++) 16 {17 for (Int J = 0; j <n; j ++) 18 {19 CD [I * n + J] = C [I] + d [J]; 20} 21} 22 sort (Cd, CD + N * n ); 23 long res = 0; 24 for (INT I = 0; I <n; I ++) 25 {26 for (Int J = 0; j <N; j ++) 27 {28 int Cd =-(A [I] + B [J]); 29 // take out part 30 res + = upper_bound (Cd, CD + N * n, Cd)-lower_bound (Cd, CD + N * n, CD); 31} 32} 33 printf ("% LLD \ n", Res); 34} 35 36 37 int main () 38 {39 CIN> N; 40 for (INT I = 0; I <n * 4; I ++) 41 {42 switch (I/6) 43 {44 case 0: 45 CIN> A [I % 6]; 46 break; 47 Case 1: 48 CIN> B [I % 6]; 49 break; 50 Case 2: 51 CIN> C [I % 6]; 52 break; 53 Case 3: 54 CIN> d [I % 6]; 55 break; 56 default: 57 break; 58} 59} 60 solve (); 61 Return 0; 62}

 

Semi-enumeration (bidirectional search)