Poj 2549 --- the legendary bucket used to prevent hash conflicts

Http://poj.org/problem? Id = 2549

Wikipedia has 3sum algorithm: https://en.wikipedia.org/wiki/3SUM

 sort(S); for i=0 to n-3 do    a = S[i];    k = i+1;    l = n-1;    while (k<l) do       b = S[k];       c = S[l];       if (a+b+c == 0) then          output a, b, c;          // Continue search for all triplet combinations summing to zero.           k = k + 1           l = l - 1       else if (a+b+c > 0) then          l = l - 1;       else          k = k + 1;       end       end end

Hash is used here:

1. sort and obtain the sum of any two numbers. Map the sum with (I, j) through the hash array, handle hash conflicts using the adjacent table in graph theory-the bucket hash discussed by the instructor when learning the Data Structure

2. enumerate the difference between two numbers (p, q). Find (I, j) through hash to ensure that all values of P, Q, I, and J are not equal.

Over 40 ms

# Include <cstdio> # include <cstring> # include <algorithm> # include <string> # include <iostream> # include <cmath> # include <map> # include <set> # include <queue> using namespace STD; # define ls (RT) rt * 2 # define RS (RT) rt * 2 + 1 # define ll long # define ull unsigned long # define rep (I, S, e) for (INT I = s; I <E; I ++) # define repe (I, S, e) for (INT I = s; I <= E; I ++) # define Cl (a, B) memset (a, B, sizeof (A) # define in (s) freopen (s), "R", stdin) # define out (s) freopen (S, "W", stdin) const int size = 536870911; const int mod = 9973; const int maxn = 1000 + 100; ll s [maxn]; struct node {int I; Int J; int next;} rec [1100005]; int hash [1100005]; // header node inline ll ABS (ll x) {return x> 0? X :(-x);} int main () {// In ("poj2549.txt"); int N, CNT; while (~ Scanf ("% d", & N) {Cl (hash, 0xff); For (INT I = 0; I <n; I ++) scanf ("% LLD", & S [I]); sort (S, S + n); CNT = 0; For (INT I = 0; I <n-1; I ++) for (Int J = I + 1; j <n; j ++) {rec [CNT]. I = I; REC [CNT]. j = J; ll sum = ABS (s [I] + s [J]) % MOD; REC [CNT]. next = hash [Sum]; hash [Sum] = CNT; CNT ++;} int OK = 0; ll ans = 0; For (INT I = n-1; I> 0; I --) {If (OK) break; For (Int J = 0; j <I; j ++) {If (OK) break; ll sum = ABS (s [I]-s [J]) % MOD; For (INT v = hash [Sum]; V! =-1; V = rec [v]. next) {int H = rec [v]. i; int K = rec [v]. j; If (s [H] + s [k] = s [I]-s [J] & rec [v]. i! = I & rec [v]. I! = J & rec [v]. J! = J & rec [v]. J! = I) {OK = 1; ans = s [I]; break ;}}} if (OK) printf ("% LLD \ n", ANS ); else printf ("no solution \ n");} return 0 ;}