poj 2785 4 Values whose Sum is 0（折半枚举（双向搜索））

Posted 2020-07-06 UniqueColor

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).

 

AC代码：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <cmath>
 6 #include <set>
 7 using namespace std;
 8 #define ll long long
 9 #define N 4006
10 int n;
11 int mp[N][N];
12 int A[N],B[N],C[N],D[N];
13 int CD[N*N];
14 void solve(){
15    for(int i=0;i<n;i++){
16        for(int j=0;j<n;j++){
17           CD[i*n+j]=C[i]+D[j];
18        }
19    }
20    sort(CD,CD+n*n);
21    ll ans=0;
22    for(int i=0;i<n;i++){
23        for(int j=0;j<n;j++){
24           int cd = -(A[i]+B[j]);
25           ans+=upper_bound(CD,CD+n*n,cd)-lower_bound(CD,CD+n*n,cd);
26        }
27    }
28    printf("%I64d\n",ans);
29 }
30 int main()
31 {
32     while(scanf("%d",&n)==1){
33        for(int i=0;i<n;i++){
34           for(int j=0;j<4;j++){
35              scanf("%d",&mp[i][j]);
36           }
37        }
38        for(int i=0;i<n;i++){
39           A[i] = mp[i][0];
40           B[i] = mp[i][1];
41           C[i] = mp[i][2];
42           D[i] = mp[i][3];
43        }
44        solve();
45     }
46     return 0;
47 }