Uva1152 4 Values whose Sum is 0

Uva1152 4 Values whose Sum is 0

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 2
28) 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).

Solution

这道题若是我们直接枚举，时间复杂度达到O(n^4)显然不可行。因此我们可以考虑折半枚举(中途相遇法)，即：每次先用hash表记录前两个集合中ai+bj的和，然后用hash表判断后两个集合中是否存在ck+dj=-ai-bj

• 代码

#include <vector>#include <cstdio>#include <cstdlib>using namespace std;const int Inf = 2147483647;const int maxn = 4005, maxl = 100987, maxs = 4001 * 4001;vector <int> Hash[maxl];int s[5][maxn], n, t, ans;inline int read(){    char c;    do {      c = getchar();    }while(c < '0' || c > '9');    int sum = 0;    do {      sum = sum * 10 + c - 48;      c = getchar();    }while(c >= '0' && c <= '9');    return sum;}inline int abs(int a){    if(a > 0) return a;    return -a;}inline void push(int x){    Hash[abs(x) % maxl].push_back(x);}inline int search(int x){    int i, k = abs(x) % maxl;    int sz = Hash[k].size(), sum = 0;    for(i = 0; i < sz; i++)      if(Hash[k][i] == x) sum++;    return sum;}int main(){    freopen("input.in", "r", stdin);    freopen("output.out", "w", stdout);    register int i, j;    scanf("%d", &t);    for(int k = 1; k <= t; k++) {      for(i = 0; i < maxl; i++)  //记得将hash表清0        Hash[i].clear();      scanf("%d", &n);      for(j = 1; j <= n; j++)        for(i = 1; i <= 4; i++)          scanf("%d", &s[i][j]);      for(i = 1; i <= n; i++)        for(j = 1; j <= n; j++)          push(s[1][i] + s[2][j]);      ans = 0;      for(i = 1; i <= n; i++)        for(j = 1; j <= n; j++)          ans += search(-s[3][i] - s[4][j]);      printf("%d\n", ans);      if(k < t) printf("\n");     }    return 0;}