POJ 2785 4 Values whose Sum is 0
4 Values whose Sum is 0
Time Limit: 15000MS | Memory Limit: 228000K | |
Total Submissions: 20015 | Accepted: 5974 | |
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 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
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
解析:折半枚举。直接从4个数列中选择的话有n4种情况,时间复杂度太高,不可取。可以采取折半枚举的方式,将A、B、 C、 D分为AB和CD考虑,从A、B中取出a、b后,为了使总和为0则需要从C、D中取出c+d = -(a+b)。所以先将C、D取数的n2种方法枚举出来,并排好序,这样就可以运用二分搜索了,时间复杂度为O(n2logn2)。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 | #include <cstdio> #include <algorithm> #include <vector> #define ll long long using namespace std; const int MAXN = 4000+5; int a[MAXN], b[MAXN], c[MAXN], d[MAXN]; int cd[MAXN*MAXN]; int n; pair<vector< int >::iterator, vector< int >::iterator> it; void solve() { for ( int i = 0; i < n; ++i){ for ( int j = 0; j < n; ++j){ cd[i*n+j] = c[i]+d[j]; } } sort(cd, cd+n*n); ll res = 0; for ( int i = 0; i < n; ++i){ for ( int j = 0; j < n; ++j){ int c_d = -(a[i]+b[j]); it = equal_range(cd, cd+n*n, c_d); res += it.second-it.first; } } printf ( "%I64d\n" , res); } int main() { scanf ( "%d" , &n); for ( int i = 0; i < n; ++i) scanf ( "%d%d%d%d" , &a[i], &b[i], &c[i], &d[i]); solve(); return 0; } |
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