Submission #591457

#	Time	Username	Problem	Language	Result	Execution time	Memory
591457		zaneyu	City Mapping (NOI18_citymapping)	C++14	100 / 100	6 ms	596 KiB

citymapping

/*input
4 200 3
2 3 1
3 4 2
1 2 9
*/
#include "citymapping.h"
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef tree<long long,null_type,less_equal<long long>,rb_tree_tag,tree_order_statistics_node_update> indexed_set;
#pragma GCC optimize("O2","unroll-loops","no-stack-protector")
//order_of_key #of elements less than x
// find_by_order kth element
using ll=long long;
using ld=long double;
using pii=pair<ll,ll>;
#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()))))
const ll INF64=4e18;
const int INF=0x3f3f3f3f;
const ll MOD=1e9+7;
const ld PI=acos(-1);
const ld eps=1e-9;
#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)
inline ll mult(ll a,ll b){
    if(a<0) a+=MOD;
    if(b<0) b+=MOD;
    if(a>=MOD) a%=MOD;
    if(b>=MOD) b%=MOD;
    return (a*b)%MOD;
}
inline ll mypow(ll a,ll b){
    if(b<=0) return 1;
    ll res=1LL;
    while(b){
        if(b&1) res=mult(res,a);
        a=mult(a,a);
        b>>=1;
    }
    return res;
}
const int maxn=1e3+5;
ll dep[maxn];
int sz[maxn];
vector<pii> v[maxn];
bool vis[maxn];
int par[maxn];
int dfs(int u,int p){
    int mx=0,vv=-1;
    for(auto x:v[u]){
        if(x.f!=p and !vis[x.f]){
            if(sz[x.f]>mx){
                mx=sz[x.f],vv=x.f;
            }
        }
    }
    if(vv==-1) return u;
    return dfs(vv,u);
}
void connect(int a,int b){
    //cout<<a<<' '<<b<<'\n';
    par[b]=a;
    v[a].pb({b,dep[b]-dep[a]});
    sz[b]=1;
    sz[1]++;
    while(a!=1){
        sz[a]++;
        a=par[a];
    }
}
void find_roads(int n, int q, int A[], int B[], int W[]) {
    vector<pii> vv;
    par[1]=-1;
    for(int i=2;i<=n;i++){
        dep[i]=get_distance(1,i);
        vv.pb({dep[i],i});
    }
    sort(ALL(vv));
    sz[1]=1;
    REP(i,sz(vv)){
        REP1(j,n) vis[j]=0; 
        int z=vv[i].s,rt=1;
        while(true){
            int p=dfs(rt,par[rt]);
            if(p==rt){
                connect(p,z);
                break;
            }
            ll a=dep[p]-dep[rt],b=dep[z]-dep[rt],c=get_distance(z,p);
            if(b==a+c){
                connect(p,z);
                break;
            }
            ll tot=(a+b-c)/2;
            tot+=dep[rt];
            while(dep[p]>tot) vis[p]=1,p=par[p];
            rt=p;
        }
    }
    int cnt=0;
    REP1(i,n){
        for(auto x:v[i]){
            A[cnt]=i,B[cnt]=x.f,W[cnt]=x.s;
            ++cnt;
        }
    }
}

• Subtask 1: 9 / 9
• Subtask 2: 16 / 16
• Subtask 3: 13 / 13
• Subtask 4: 19 / 19
• Subtask 5: 43 / 43