It is already well-known that using unordered series in Codeforces is super dangerous, and it is not recommended to use them at all unless you know how to prevent them properly (well, even if you know, just using a regular map or set is more beneficial in most cases). Thanks to ...
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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++) ...
#include<bits/stdc++.h>usingnamespacestd;constintN=2e5+10,INF=2e9+10;#define int long longintn,t,x,y,m,q[N],p[N];strings;voidsolve(){cin>>n;unordered_map<string,int>mp;vector<string>s1;for(inti=1;i<=n;i++){cin>>s;s1.push_back(s);mp[s]++;}for(inti=0;i<s1.size...
m, like[maxn], w[maxn];intsum[2], cnt[2], val[maxn * maxn][maxn * maxn], dp[maxn][maxn <<1][5000][2];//in fact, 2500 is okunordered_map<longlong,int> mp;inthashcnt;intksm(inta,intb){intres =1;for(; b; b >>=1) {if(b &1) res =1LL* res * a % mod...
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++){ ...
1#include <iostream>2#include <unordered_map>3#include <algorithm>4#include <vector>5#include <queue>67usingnamespacestd;89boolsppalindjudge(strings,intn)10{11char* pre = &s[0];12char* end = &s[n-1];13while(pre <end)14{15if(abs((*pre)-(*end))==0|| abs((*pre)-(*end))...
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; } }