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http://acm.hdu.edu.cn/showproblem.php?pid=4334
Hassan is in trouble. His mathematics teacher has given him a very difficult problem called 5-sum. Please help him.
The 5-sum problem is defined as follows: Given 5 sets S_1,...,S_5 of n integer numbers each, is there a_1 in S_1,...,a_5 in S_5 such that a_1+...+a_5=0?
First line of input contains a single integer N (1≤N≤50). N test-cases follow. First line of each test-case contains a single integer n (1<=n<=200). 5 lines follow each containing n integer numbers in range [-10^15, 1 0^15]. I-th line denotes set S_i for 1<=i<=5.
For each test-case output "Yes" (without quotes) if there are a_1 in S_1,...,a_5 in S_5 such that a_1+...+a_5=0, otherwise output "No".
2
2
1 -1
1 -1
1 -1
1 -1
1 -1
3
1 2 3
-1 -2 -3
4 5 6
-1 3 2
-4 -10 -1
No
Yes
简单哈希。。
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<map>
using std::abs;
using std::min;
using std::find;
using std::pair;
using std::vector;
using std::multimap;
#define pb(e) push_back(e)
#define sz(c) (int)(c).size()
#define mp(a, b) make_pair(a, b)
#define all(c) (c).begin(), (c).end()
#define iter(c) __typeof((c).begin())
#define cls(arr, val) memset(arr, val, sizeof(arr))
#define cpresent(c, e) (find(all(c), (e)) != (c).end())
#define rep(i, n) for(int i = 0; i < (int)n; i++)
#define tr(c, i) for(iter(c) i = (c).begin(); i != (c).end(); ++i)
const int N = 1000007;
const int INF = 0x3f3f3f3f;
typedef long long ll;
struct Hash_Set {
ll num[N << 1];
int tot, head[N], next[N];
inline void init() {
tot = 0, cls(head, -1);
}
inline void insert(ll val) {
ll u = abs(val) % N;
num[tot] = val, next[tot] = head[u], head[u] = tot++;
}
inline bool find(ll val) {
ll u = abs(val) % N;
for(int i = head[u]; ~i; i = next[i]) {
if(num[i] == val) return true;
}
return false;
}
}hash;
ll A[5][210];
int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w+", stdout);
#endif
ll val;
int t, n;
scanf("%d", &t);
while(t--) {
hash.init();
scanf("%d", &n);
rep(i, 5) {
rep(j, n) scanf("%lld", &A[i][j]);
}
rep(i, n) {
rep(j, n) {
val = A[0][i] + A[1][j];
hash.insert(val);
}
}
bool f = false;
rep(i, n) {
rep(j, n) {
rep(k, n) {
val = A[2][i] + A[3][j] + A[4][k];
val = -val;
if(hash.find(val)) { f = true; break; }
}
if(f) break;
}
if(f) break;
}
puts(f ? "Yes" : "No");
}
return 0;
}
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原文地址：http://www.cnblogs.com/GadyPu/p/4792966.html
