HDU 2141 哈希表或者二分

题目大意：

Ai + Bj + Ck == X，是否能成立

思路：

因为是l*n*m = 500 * 500 * 500，还有s的1000，如果暴力枚举肯定是超时。我之前思考过用vector动态处理内存，然后用erase和unique+sort将里面所有的重复的元素消除，不过好像超时了。

然后就想到将前两项的和加在一起，在用二分法判断

或者用哈希表，将所有的元素都一个一个的表示出来，然后在取判断即可。

哈希表是400+MS，二分法是900+MS。主要原因在于，哈希表的判断是接近O（1），而二分法是一个log250000，然后1000*500的基数较大，所以差别较大

哈希表的

#include<cstdio>#include<algorithm>#include<cstring>using namespace std;typedef long long ll;const int mod = 300007;const int maxn = 500 + 10;const int maxm = 300007;ll l, n, m, s, tmp, cnt;ll a[maxn], b[maxn], c[maxn];ll to[250000 + 5], head[maxm + 1], nex[250000 + 5];void init(){ memset(a, 0, sizeof(a)); memset(b, 0, sizeof(b)); memset(c, 0, sizeof(c)); for (int i = 0; i < 250000; i++) nex[i] = to[i] = -1; for (int i = 0; i < maxm + 1; i++) head[i] = -1; cnt = 0; for (int i = 0; i < l; i++) scanf("%I64d", a + i); for (int i = 0; i < n; i++) scanf("%I64d", b + i); for (int i = 0; i < m; i++) scanf("%I64d", c + i);}void add(ll val){ ll ret = (val % mod + mod) % mod; for (ll i = head[ret]; i != -1; i = nex[i]){ if (val == to[i]) return ; } to[cnt] = val; nex[cnt] = head[ret]; head[ret] = cnt++; return ;}bool C(ll x){ ll ret = (x % mod + mod) % mod; for (ll i = head[ret]; i != -1; i = nex[i]){ if (to[i] == x) return true; } return false;}void solve(){ for (int i = 0; i < l; i++){ for (int j = 0; j < n; j++){ ll sum = a[i] + b[j]; add(sum); } } ll it; while (s--){ scanf("%I64d", &it); bool flag = 0; for (int i = 0; i < m; i++){ if (C(it - c[i])){ flag = 1; break; } } if (flag) printf("YES\n"); else printf("NO\n"); }}int main(){ while (scanf("%I64d%I64d%I64d", &l, &n, &m) == 3){ init(); scanf("%I64d", &s); printf("Case %I64d:\n", ++tmp); solve(); } return 0;}</cstring></algorithm></cstdio>

 二分法：

将数组组合起来。估算一下复杂度即可，判断最大为O（500 * 500 + 1000 * 500 * log250000），log250000可以看做是个位数的即可。

#include<cstdio>#include<algorithm>#include<cstring>using namespace std;typedef long long ll;const int maxn = 500 + 5;ll cnt, l, n, m, s, tmp;ll a[maxn], b[maxn], c[maxn];ll sum[maxn * maxn];void init(){ memset(a, 0, sizeof(a)); for (int i = 0; i < l; i++){ scanf("%I64d", a + i); } memset(b, 0, sizeof(b)); for (int i = 0; i < n; i++){ scanf("%I64d", b + i); } memset(c, 0, sizeof(c)); for (int i = 0; i < m; i++){ scanf("%I64d", c + i); } for (int i = 0; i < maxn * maxn; i++) sum[i] = 0; tmp = 0; for (int i = 0; i < l; i++){ for (int j = 0; j < n; j++){ sum[tmp++] = a[i] + b[j]; } } sort(sum, sum + tmp - 1);}bool C(ll x){ int lb = 0, ub = tmp - 1; while (ub - lb > 1){ ll mid = (lb + ub) / 2; if (sum[mid] == x || sum[mid + 1] == x) return true; if (sum[mid] < x) lb = mid; else ub = mid; } return false;}void solve(){ printf("Case %d:\n", ++cnt); while (s--){ ll x; scanf("%I64d", &x); bool flag = 0; for (int i = 0; i < m; i++){ if (C(x - c[i])){ flag = 1; break; } } if (flag) printf("YES\n"); else printf("NO\n"); }}int main(){ while (scanf("%I64d%I64d%I64d", &l, &n, &m) == 3){ init(); scanf("%I64d", &s); solve(); } return 0;}</cstring></algorithm></cstdio>