POJ 2785 (二分)

4 Values whose Sum is 0 Time Limit: 15000MS   Memory Limit: 228000K Total Submissions: 18785   Accepted: 5579 Case Time Limit: 5000MS 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 2 28 ) 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). Source Southwestern Europe 2005

题意是给出4个数列，每个数列n个数，要求从每个数列中取出一个数使得4个数的和为0。

先取两组记录两两的和，然后枚举另外两组两两的和从已知的和里面去找。

可能map常数比较大过不了，二分就可以了。

#include <iostream>
#include <cmath>
#include <cstdlib>
#include <time.h>
#include <vector>
#include <cstdio>
#include <cstring>
#include <map>
#include <algorithm>
using namespace std;
#define maxn 4111

int n;
long long a[maxn][4];
long long sum[maxn*maxn];

int main () {
	while (scanf ("%d", &n) == 1) {
		for (int i = 1; i <= n; i++) {
			scanf ("%lld%lld%lld%lld", &a[i][0], &a[i][1], &a[i][2], &a[i][3]);
		}
		int cnt = 0;
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= n; j++) {
				long long num = a[i][2]+a[j][3];
				sum[cnt++] = num;
			}
		}
		long long ans = 0;
		sort (sum, sum+cnt);
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= n; j++) {
				long long num = -1 * (a[i][0] + a[j][1]);
				ans += upper_bound (sum, sum+cnt, num) - lower_bound (sum, sum+cnt, num);
			}
		}
		printf ("%lld\n", ans);
	}
	return 0;
}