Submission #638995

#	Time	Username	Problem	Language	Result	Execution time	Memory
638995		zaneyu	Traffic (IOI10_traffic)	C++14	100 / 100	890 ms	170716 KiB

traffic

/*input
3
aa
acb
c
*/
#include<bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<long long,null_type,less_equal<long long>,rb_tree_tag,tree_order_statistics_node_update> indexed_set;
//order_of_key #of elements less than x
// find_by_order kth element
using ll=long long;
using ld=long double;
using pii=pair<int,int>;
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(int i=0;i<n;i++)
#define REP1(i,n) for(int i=1;i<=n;i++)
#define FILL(n,x) memset(n,x,sizeof(n))
#define ALL(_a) _a.begin(),_a.end()
#define sz(x) (int)x.size()
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()),c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define GET(v,x) lower_bound(ALL(v),x)-v.begin()
const int maxn=1e6+3;
const int INF=0x3f3f3f3f;
const int MOD=1e6+3;
const ld PI=acos(-1.0l);
const ld eps=1e-6;
const ll INF64=9e18+1;
#define lowb(x) x&(-x)
#define MNTO(x,y) x=min(x,(__typeof__(x))y)
#define MXTO(x,y) x=max(x,(__typeof__(x))y)
template<typename T1,typename T2>
ostream& operator<<(ostream& out,pair<T1,T2> P){
    out<<P.f<<' '<<P.s;
    return out;
}
template<typename T>
ostream& operator<<(ostream& out,vector<T> V){
    REP(i,sz(V)) out<<V[i]<<((i!=sz(V)-1)?"\n":"");
    return out;
}
int mult(int a,int b){
    return 1LL*a*b%MOD;
}
ll mypow(ll a,ll b,ll mod){
    if(b<=0) return 1;
    a%=mod;
    ll res=1LL;
    while(b){
        if(b&1) res=(res*a)%mod;
        a=(a*a)%mod;
        b>>=1;
    }
    return res;
}
int n;
ll sz[maxn],tot=0;
ll ans=INF64;
int ppp;
vector<int> v[maxn];
void dfs(int u,int p){
    ll mx=0;
    for(int x:v[u]){
        if(x==p) continue;
        dfs(x,u);
        sz[u]+=sz[x];
        MXTO(mx,sz[x]);
    }
    MXTO(mx,tot-sz[u]);
    MNTO(ans,mx);
    if(mx==ans) ppp=u;
}
int LocateCentre(int N, int pp[], int S[], int D[]) {
    n=N;
    REP(i,n){
        sz[i]=pp[i];
        tot+=pp[i];
    }
    REP(i,n-1) v[S[i]].pb(D[i]),v[D[i]].pb(S[i]);
    dfs(0,-1);
    return ppp;
}

• Subtask 1: 25 / 25
• Subtask 2: 25 / 25
• Subtask 3: 25 / 25
• Subtask 4: 25 / 25