poj 2785 4 Values whose Sum is 0

题目连接

http://poj.org/problem?id=2785  

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
4
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
5
-56 45 34 78
-34 -67 45 -23
-12 -34 -56 46
45 34 -32 8
-23 -56 4 53

Sample Output

5
256
11

哈希大法。。

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<set>
using std::abs;
using std::sort;
using std::pair;
using std::swap;
using std::queue;
using std::multiset;
#define pb(e) push_back(e)
#define sz(c) (int)(c).size()
#define mp(a, b) make_pair(a, b)
#define all(c) (c).begin(), (c).end()
#define iter(c) decltype((c).begin())
#define cls(arr, val) memset(arr, val, sizeof(arr))
#define cpresent(c, e) (find(all(c), (e)) != (c).end())
#define rep(i, n) for(int i = 0; i < (int)n; i++)
#define tr(c, i) for(iter(c) i = (c).begin(); i != (c).end(); ++i)
const int N = 5010;
const int M = 3000007;
const int INF = 0x3f3f3f3f;
typedef long long ll;
ll A[4][N];
struct HashSet {
	struct Node {
		ll v;
		int cnt;
	}arr[N * N];
	int tot, head[M], next[M];
	inline void init() {
		tot = 0;
		rep(i, M) head[i] = -1;
	}
	inline void insert(ll val) {
		bool f = false;
		int u = abs(val) % M;
		for (int i = head[u]; ~i; i = next[i]) {
			if (arr[i].v == val) {
				f = true;
				arr[i].cnt++;
			}
		}
		if (!f) {
			arr[tot].v = val, arr[tot].cnt = 1; next[tot] = head[u], head[u] = tot++;
		}
	}
	inline int find(ll val) {
		int u = abs(val) % M;
		for (int i = head[u]; ~i; i = next[i]) {
			if (arr[i].v == val) return arr[i].cnt;
		}
		return 0;
	}
}work;
int main() {
#ifdef LOCAL
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w+", stdout);
#endif
	int n;
	while (~scanf("%d", &n)) {
		ll ans = 0;
		work.init();
		rep(i, n) {
			rep(j, 4) scanf("%lld", &A[j][i]);
		}
		rep(i, n) {
			rep(j, n) {
				work.insert(A[0][i] + A[1][j]);
			}
 		}
		rep(i, n) {
			rep(j, n) {
				ans += work.find(-(A[2][i] + A[3][j]));
			}
		}
		printf("%lld
", ans);
	}
	return 0;
}