UVA - 1152（unordered_map的应用）

4 Values whose Sum is 0 UVA - 1152
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 × B × C × D are such that a + b + c + d = 0. In the following, we
assume that all lists have the same size n.
Input
The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.
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 test case, your program has to write the number quadruplets whose sum is zero.
The outputs of two consecutive cases will be separated by a blank line.
Sample Input
1
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
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)
思路：
遍历A，B，将a+b的结果存到哈希表中；
再遍历C，D，ans+=mp[-(C[i]+D[j])]；
用map时间复杂度为：O（n^2*logn）
用unordered_map时间复杂度：O（n^2）
相较于蛮力算法O（n^4）已经优化了很多了！
注意点：
1.这个题用map是超时的，需要用到unordered_map；
2.注意回车控制！

ac code：

#include
using namespace std;
typedef long long ll;
const int maxn=5000;
int A[maxn],B[maxn],C[maxn],D[maxn];
unordered_map<int,int> mp;//unordered_map
ll ans;
int main(){
     
	int t,n;
	scanf("%d",&t);
	int first=1;
	while(t--){
     
        if(first){
     
            first=0;
        }else{
     
            printf("\n");
        }
		ans=0;
		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++){
     
				mp[A[i]+B[j]]++;
			}
		}

		for(int i=0;i<n;i++){
     
			for(int j=0;j<n;j++){
     
				ans+=mp[-C[i]-D[j]];
			}
		}
		printf("%lld\n",ans);
		mp.clear();
	}
	return 0;
}