Using unordered map with custom hash mentioned below for storing frquency/kind of prefix sum but gives TLE and using map gets accepted. For the first time I have seen unordered map with this custom hash is getting TLE because of it. (Custom hash fromblog) structcustom_hash{staticuint64_ts...
An unordered_map is a hash table. Hash tables can't be looked at in any meaningful order. They work by running what is called a hash function on inserted values to generate a number, which is used to index the stored value. They can be added to and accessed in O(1) time. The com...
但unordered_map用的是传统哈希函数构造,这意味着可能会发生哈希冲突。当 n 较小时,发生冲突的概率很小,时间复杂度可以看作 O(1) ;但当 n 比较大,会有概率发生冲突,而最坏可能和之前的全部冲突,时间复杂度 O(n)。 代码 #include<bits/stdc++.h> #define ll long long #define U unsigned #define sqr(...
在STL 中,还有 multiset 与 unordered_set。 multiset,就是允许重复元素的集合(你甚至可以把其理解为平衡树)。 unordered_set,就是不允许重复元素,但是不排序的集合。 2-2-2. STL map, multimap, unordered_map map,即映射,将一个值映射到另一个值。map 内部同样是一棵红黑树。 定义一个 map: map<int,in...
#include<map> #include<queue> #include<unordered_map> #include<cmath> #include<cstdio> #include<algorithm> #include<set> #include<cstdlib> #include<stack> #define forxn(i,a,n) for(int i=a;i<n;i++) #define fordn(i,a,n) for(int i=a;i<=n;i++) ...
每个分块维护一个 unordered_map 记录块中元素的出现次数, 查询时从前往后找第一个包含目标值的分块, 从后往前找第一个包含目标值的分块, 途中跳过不含目标值的分块即可. 总时间复杂度 O(qn) . 代码 struct myHash { static uint64_t splitmix64(uint64_t x) { x += 0x9e3779b97f4a7c15; x =...
于是我们可以用unordered_map记录所有存在的度数,枚举n的因数x,y,若度数x和度数y + 1存在或者度数y和度数x + 1存在,则x和y就是满足要求的解。code:#include <bits/stdc++.h> using namespace std; #define x first #define y second typedef pair<int, int> PII; typedef long long LL; const int N...
unordered_map<ull,int>mp; ullget(intl,intr){if(l==0)returnHash[r];returnHash[r]-Hash[l-1]*p[r-l+1]; }intL[N],id[N],a[N]; ull H[N];boolcmp(intx,inty){returnL[x]<L[y]; } vector<int>ans[N];intmain(){ fio;stringstr; ...
unordered_map<double ,set<double> > ma; int n; int main() { scanf("%d", &n); for(int i=0;i<n;i++){ double x, y; scanf("%lf%lf",&x, &y); v.push_back(p(x, y)); } for(int i=0;i<v.size();i++){ for(int j=i+1;j<v.size();j++){ ...
unordered_map<int, vector<int>> mp; int dfs(int val, int p, int w){ int t = a[val]; for(int x: mp[val]){ if (x != p) { int k = dfs(x, val, w); if (k == w) { nums ++; } else t = t ^ k; } }