解题报告 之 UVA1152 4 Values Whose Sum is Zero

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 ) AxBxCxD 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, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.

For each input file, your program has to write the number quadruplets whose sum is zero.

Sample Input 

16-45 22 42 -16-41 -27 56 30-36 53 -37 77-36 30 -75 -4626 -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).

题目大意：给出4个数据集A,B,C,D，从其中各取一个数字，问取出的数字相加等于0有多少种取法？

好吧，不得不承认一开始用Hash表来弄时间跪了。后来看了一下分析是中途相遇法。最朴素的想法是枚举A,B,C,D。复杂度O（n^4），只能呵呵哒。然后再想想发现可以枚举A,B,C，再看与D中有没有其相反数。然而复杂度仍然是O（n^3*logn）——枚举+查找。所以顺着这样想，就是中间相遇法，即枚举A+B和C+D，再在C+D中去查找A+B的相反数。那么复杂度就是O（n^2*logn）。

先按照紫书说的用hash结果时间挂掉了，估计是查找没有优化还是O（n^3）。于是乎看了一下题解。发现了一个高端的方法，先将A+B排序，在使用STL，upper_bound(a,a+n,key)和lower_bound(a,a+n,key)。upper_bound是返回a~a+n之间第一个大于key的迭代器而lower_bound是返回第一个大于等于的迭代器，则两个结果相减就是key的个数，那么我们这道题就是用upper_bound（aab，aab+cnt，-C[i]-D[j]）- lower_bound（aab，aab+cnt，-C[i]-D[j]）就可以了。

下面上代码：

#include <iostream>#include <cmath>#include <algorithm>#define MAXN 4010using namespace std;long long  a[MAXN], b[MAXN], c[MAXN], d[MAXN];long long  aab[MAXN*MAXN];int main(){int kase, n;cin >> kase;while (kase--){cin >> n;for (int i = 0; i < n; i++){cin >> a[i] >> b[i] >> c[i] >> d[i];}long long cnt = 0;for (int i = 0; i < n; i++){for (int j = 0; j < n; j++){aab[cnt++] = a[i] + b[j];}}sort(aab, aab + cnt);int ans = 0;for (int i = 0; i < n; i++){for (int j = 0; j < n; j++){ans += upper_bound(aab, aab + cnt, -c[i] - d[j]) - lower_bound(aab, aab + cnt, -c[i] - d[j]);}}cout << ans << endl;if (kase) cout << endl;}return 0;}

另外有几道中途相遇法的题，先列在这以后做吧，毕竟最近再赶任务。

 

hdu 4963 Dividing a String

 

uva 1326 Jurassic Remains

 POJ1903 - Jurassic Remains