PKU Acm2785--4 Values whose Sum is 0~~ bidirectional search

The meaning of the topic is very clear, ABCD four sets, extracting a number from each set, and adding four numbers equals 0 of the combined number.

Because n maximum reaches 4000, so, the method of violence can not live, is the time limit is 15000MS also not.

But we can search two-way, that is binary search.

First find the CD of the two sets of 22 pairs of new collection CD, and then in the collection from the CD to look for AB set 22 paired and K negative number of how many. The Upper_bound and Lower_bound functions can be solved quickly.

To put it simply, a+b+c+d=0, calculate the C + d,a + b =-(c + D) and look it up in the collection CD.

The following is the code for the AC:

#include <iostream> #include <algorithm> #include <cstdio>using namespace Std;int n;int a[4005], b[ 4005], c[4005], d[4005];int cd[4005 * 4005];void solve () {int I, J, k;for (i = 0; i < n; i++) for (j = 0; J < N; j + +) cd[ I * n + j] = C[i] + d[j];        For collection Cdsort (CD, CD + n * n);//Sort __int64 ans = 0;for (i = 0; i < n; i++) {for (j = 0; J < N; j + +) {k = A[i] + B[j];ans + = Upper_bound (CD, CD + N * N,-K)-Lower_bound (CD, CD + N * N,-K);   Find out how many elements are equal to-K}}printf ("%i64d\n", ans);} int main () {int i;scanf ("%d", &n), for (i = 0; i < n; i++) {scanf ("%d", &a[i]), scanf ("%d", &b[i]), scanf ("%d", and Amp C[i]); scanf ("%d", &d[i]);} Solve (); return 0;}

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PKU Acm2785--4 Values whose Sum is 0~~ bidirectional search