(Hackerrank) Tower Breakers, Again!

https://www.hackerrank.com/challenges/kitty-and-katty

#include <iostream>
#include <cstdio>
#include <vector>
#include <utility>
#include <set>
#include <queue>
#include <cstring>
#include <algorithm>
using namespace std;

#define SZ(x) ((int)(x).size())
typedef pair<int,int> pii;
const int MAX_N = 1e5 +6;
int prime[MAX_N];

void build() {
    prime[0] = prime[1] = 1;
    for (int i=2;MAX_N>i;i++) {
        if (prime[i] == 0) {
            prime[i]=i;
            for (long long j=i*1LL*i;MAX_N>j;j+=i) {
                prime[j] = i;
            }
        }
    }
}

vector<pii> get_prime_divisor(int x) {
    vector<pii> ret;
    while (x!=1) {
        pii tmp;
        tmp.first = prime[x];
        int cnt=0;
        int pp=prime[x];
        while (x%pp==0) {
            x/=pp;
            cnt++;
        }
        tmp.second = cnt;
        ret.emplace_back(tmp);
    }
    return ret;
}

vector<int> get_divisor(int x) {
    vector<pii> prime_divisor=get_prime_divisor(x);
    int sz=SZ(prime_divisor);
    queue<pii> que;
    que.push({1,0});
    vector<int> ret;
    while (!que.empty()) {
        pii p=que.front();
        que.pop();
        if (p.second == sz) {
            ret.emplace_back(p.first);
            continue;
        }
        int t=1;
        for (int i=0;prime_divisor[p.second].second>=i;i++) {
            que.push(make_pair(p.first*t,p.second+1));
            t*=prime_divisor[p.second].first;
        }
    }
    return ret;
}

int dp[MAX_N];

const int MAGIC = 60;

void doo(int id) {
    vector<int> ret=get_divisor(id);
    int cnt[MAGIC];
    sort(ret.begin(),ret.end());
    memset(cnt,0,sizeof(cnt));
    for (int i:ret) {
        if (i==id) continue;
        if (dp[i] == -1) doo(i);
        cnt[((id/i)%2==0?0:dp[i])]=1;
    }
    for (int i=0;MAGIC>i;i++) {
        if (!cnt[i]) {
            dp[id] = i;
            break;
        }
    }
}

int main () {
    build();
    int T;
    scanf("%d",&T);
    memset(dp,-1,sizeof(dp));
    dp[1] = 0;
    while (T--) {
        int n;
        scanf("%d",&n);
        int ans=0;
        for (int i=1;n>=i;i++) {
            int h;
            scanf("%d",&h);
            if (dp[h] != -1) ans ^= dp[h];
            else {
                doo(h);
                ans ^= dp[h];
            }
        }
        if (ans) puts("1");
        else puts("2");
    }
}

(Hackerrank) Tower Breakers, Revisited!

https://www.hackerrank.com/challenges/tower-breakers-revisited-1

SG value

#include <iostream>
#include <cstdio>
#include <vector>
#include <utility>
#include <set>
#include <queue>
#include <cstring>
using namespace std;

#define SZ(x) ((int)(x).size())
typedef pair<int,int> pii;
const int MAX_N = 1e6 +6;
int prime[MAX_N];

void build() {
    prime[0] = prime[1] = 1;
    for (int i=2;MAX_N>i;i++) {
        if (prime[i] == 0) {
            prime[i]=i;
            for (long long j=i*1LL*i;MAX_N>j;j+=i) {
                prime[j] = i;
            }
        }
    }
}

vector<pii> get_prime_divisor(int x) {
    vector<pii> ret;
    while (x!=1) {
        pii tmp;
        tmp.first = prime[x];
        int cnt=0;
        int pp=prime[x];
        while (x%pp==0) {
            x/=pp;
            cnt++;
        }
        tmp.second = cnt;
        ret.emplace_back(tmp);
    }
    return ret;
}

vector<int> get_divisor(int x) {
    vector<pii> prime_divisor=get_prime_divisor(x);
    int sz=SZ(prime_divisor);
    queue<pii> que;
    que.push({1,0});
    vector<int> ret;
    while (!que.empty()) {
        pii p=que.front();
        que.pop();
        if (p.second == sz) {
            ret.emplace_back(p.first);
            continue;
        }
        int t=1;
        for (int i=0;prime_divisor[p.second].second>=i;i++) {
            que.push(make_pair(p.first*t,p.second+1));
            t*=prime_divisor[p.second].first;
        }
    }
    return ret;
}

int dp[MAX_N];

const int MAGIC = 30;

void doo(int id) {
    vector<int> ret=get_divisor(id);
    int cnt[MAGIC];
    memset(cnt,0,sizeof(cnt));
    for (int i:ret) {
        if (i==id) continue;
        if (dp[i] == -1) doo(i);
        cnt[dp[i]]=1;
    }
    for (int i=0;MAGIC>i;i++) {
        if (!cnt[i]) {
            dp[id] = i;
            break;
        }
    }
}

int main () {
    build();
    int T;
    scanf("%d",&T);
    memset(dp,-1,sizeof(dp));
    dp[1] = 0;
    while (T--) {
        int n;
        scanf("%d",&n);
        int ans=0;
        for (int i=1;n>=i;i++) {
            int h;
            scanf("%d",&h);
            if (dp[h] != -1) ans ^= dp[h];
            else {
                doo(h);
                ans ^= dp[h];
            }
        }
        if (ans) puts("1");
        else puts("2");
    }
}

(Hackerrank) Balanced Forest

https://www.hackerrank.com/challenges/balanced-forest/problem

Treap + 啟發式合併

#include <iostream>
#include <cstdio>
#include <vector>
#include <ctime>
#include <cstdlib>
#include <algorithm>
using namespace std;

typedef long long LL;
const LL INF = 1e17 + 6;
const int MAX_P = 1e6 + 6;

struct Treap {
    static Treap mem[MAX_P];
    Treap *lc,*rc;
    LL key;
    int pri,sz;
    Treap(){
        
    }
    Treap(LL _key) : key(_key),pri(rand()),sz(1),lc(NULL),rc(NULL) {
        
    }
} Treap::mem[MAX_P],*ptr=Treap::mem;

int Sz(const Treap* t) {
    return t?t->sz:0;
}

void pull(Treap* t) {
    if (!t) return;
    t->sz = Sz(t->lc) + Sz(t->rc) + 1;
}

Treap* merge(Treap* a,Treap* b) {
    if (!a || !b) return a?a:b;
    else if (a->pri > b->pri) {
        a->rc=merge(a->rc,b);
        pull(a);
        return a;
    }
    else {
        b->lc=merge(a,b->lc);
        pull(b);
        return b;
    }
}

void split(Treap* t,LL k,Treap* &a,Treap* &b) {
    if (!t) a=b=NULL;
    else if (t->key <= k) {
        a=t;
        split(t->rc,k,a->rc,b);
        pull(a);
    }
    else {
        b=t;
        split(t->lc,k,a,b->lc);
        pull(b);
    }
}

bool ffind(Treap* &t,LL x) {
    Treap *tl,*tr;
    split(t,x-1,tl,t);
    split(t,x,t,tr);
    bool ret=(Sz(t)!=0);
    t=merge(tl,merge(t,tr));
    return ret;
}

const int MAX_N = 5e4 +6;

vector<int> edg[MAX_N];
LL a[MAX_N];
LL sz[MAX_N];
LL ans;

void dfs1(int id,int par) {
    sz[id] = a[id];
    for (int i:edg[id]) {
        if (i!=par) {
            dfs1(i,id);
            sz[id] += sz[i];
        }
    }
}

void addd(Treap* &t,LL val) {
    Treap *tl;
    split(t,val,tl,t);
    t=merge(merge(tl,new(++ptr)Treap(val)),t);
}

void dell(Treap* &t,LL val) {
    Treap *tl,*tr;
    split(t,val-1,tl,t);
    split(t,val,t,tr);
    t=merge(merge(tl,t->lc),merge(t->rc,tr));
}

int n;
LL tot=0;

Treap* root;

LL get_ans(LL x,LL n_2x) {
    if (x<n_2x) return INF;
    else return x-n_2x;
}

void dfs2(int id,int par) {
    //sz[id] --> x
    //see have 2*x or tot-x
    if (ffind(root,2*sz[id])) {
        ans = min(ans,get_ans(sz[id],tot-2*sz[id]));
    }
    if (ffind(root,tot-sz[id])) {
        ans = min(ans,get_ans(sz[id],tot-2*sz[id]));
    }
    //sz[id] --> tot-2*x
    //see have tot-x
    if ((tot-sz[id])%2==0) {
        LL x=(tot-sz[id])/2;
        if (ffind(root,tot-x)) {
            ans = min(ans,get_ans(x,tot-2*x));
        }
    }
    addd(root,sz[id]);
    for (int i:edg[id]) {
        if (i!=par) {
            dfs2(i,id);
        }
    }
    dell(root,sz[id]);
}

void split_by_sz(Treap* t,int k,Treap* &a,Treap* &b) {
    if (!t) a=b=NULL;
    else if (Sz(t->lc) + 1 <= k) {
        a=t;
        split_by_sz(t->rc,k-Sz(t->lc)-1,a->rc,b);
        pull(a);
    }
    else {
        b=t;
        split_by_sz(t->lc,k,a,b->lc);
    }
}

void ddfs(Treap* &a,Treap* t) {
    LL now=t->key;
    //now --> x
    //find x or tot-2*x
    if (ffind(a,now)) {
        ans = min(ans,get_ans(now,tot-2*now));
    }
    if (ffind(a,tot-2*now)) {
        ans = min(ans,get_ans(now,tot-2*now));
    }
    //now --> tot-2*x;
    //find x
    if ((tot-now)%2 == 0) {
        LL x=(tot-now)/2;
        if (ffind(a,x)) {
            ans = min(ans,get_ans(x,tot-2*x));
        }
    }
    if (t->lc) ddfs(a,t->lc);
    if (t->rc) ddfs(a,t->rc);
}

void mmerge(Treap* &a,Treap* &b) {
    ddfs(a,b);
    while(Sz(b)) {
        Treap* t;
        split_by_sz(b,1,t,b);
        Treap *tl;
        split(a,t->key-1,a,tl);
        a=merge(merge(a,t),tl);
    }
}

Treap* sagiri[MAX_N];

void dfs3(int id,int par) {
    Treap* ret=NULL;
    vector<Treap*> v;
    for (int i:edg[id]) {
        if (i!=par) {
            dfs3(i,id);
            Treap* rett=sagiri[i];
            if (Sz(rett) > Sz(ret)) {
                swap(rett,ret);
            }
            if (rett != NULL)v.push_back(rett);
        }
    }
    for (auto i:v) {
        mmerge(ret,i);
    }
    addd(ret,sz[id]);
    sagiri[id] = ret;
}

int main () {
    int T;
    scanf("%d",&T);
    while (T--) {
        ptr=Treap::mem;
        scanf("%d",&n);
        for (int i=0;n>=i;i++) {
            edg[i].clear();
        }
        for (int i=1;n>=i;i++) {
            scanf("%lld",&a[i]);
        }
        for (int i=1;n-1>=i;i++) {
            int x,y;
            scanf("%d %d",&x,&y);
            edg[x].push_back(y);
            edg[y].push_back(x);
        }
        ans = INF;
        dfs1(1,1);
        root=NULL;
        tot=sz[1];
        dfs2(1,1);
        dfs3(1,1);
        if (ans == INF) ans=-1;
        printf("%lld\n",ans);
    }
}

(Hackerrank) Kitty's Calculations on a Tree [重心剖分]

https://www.hackerrank.com/challenges/kittys-calculations-on-a-tree

重心剖分 (centroid decomposition)

#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <utility>
using namespace std;

typedef long long LL;
typedef pair<LL,LL> pii;
const int MAX_N = 2e5 + 6;
const int MAX_P = 19;
const LL mod = 1e9 + 7;

vector<int> edg[MAX_N];
int dis[MAX_P][MAX_N];
bool visit[MAX_N];

struct Cen {
    int par;
    int depth;
    pii val_v_av;  //first --> val, second --> minus
    pii val_v;
} cen[MAX_N];

vector<int> v;
int sz[MAX_N];
int mx[MAX_N];

void dfs2(int id) {
    v.push_back(id);
    visit[id]=1;
    sz[id]=1;
    mx[id]=0;
    for (int i:edg[id]) {
        if (!visit[i]) {
            dfs2(i);
            sz[id] += sz[i];
        }
    }
}

#define SZ(x) ((int)(x).size())

int get_cen(int id) {
    v.clear();
    dfs2(id);
    int tot=SZ(v);
    int cen=-1;
    for (int i:v) {
        if (max(mx[i],tot-sz[i]) <= tot/2) {
            cen=i;
        }
        visit[i]=false;
    }
    return cen;
}

void dfs3(int id,int par,int cen_depth,int dist)  {
    dis[cen_depth][id] = dist;
    for (int i:edg[id]) {
        if (!visit[i] && i!=par) {
            dfs3(i,id,cen_depth,dist+1);
        }
    }
}

void dfs(int id,int cen_par,int cen_depth) {
    int ccen=get_cen(id);
    dfs3(ccen,ccen,cen_depth,0);
    cen[ccen]={cen_par,cen_depth,{0,0},{0,0}};
    visit[ccen]=1;
    for (int i:edg[ccen]) {
        if (!visit[i]) dfs(i,ccen,cen_depth+1);
    }
}

pii operator+(const pii &p1,const pii &p2) {
    return make_pair(p1.first+p2.first,p1.second+p2.second);
}

pii operator-(const pii &p1,const pii &p2) {
    return make_pair(p1.first-p2.first,p1.second-p2.second);
}

pii operator+=(pii &p1,const pii &p2) {
    p1 = p1 + p2;
    return p1;
}

pii operator-=(pii &p1,const pii &p2) {
    p1 = p1 - p2;
    return p1;
}

void Pure(pii &p) {
    p.first = (p.first%mod + mod) % mod;
    p.second = (p.second%mod + mod) % mod;
}

void addd(LL x) {
    LL p=x;
    while (p!=-1) {
        cen[p].val_v += {x,0};
        cen[p].val_v_av += {x*dis[cen[p].depth][x],0};
        if (cen[p].par != -1) {
            int par=cen[p].par;
            cen[p].val_v -= {0,x};
            cen[p].val_v_av -= {0,x*dis[cen[par].depth][x]};
        }
        Pure(cen[p].val_v);
        Pure(cen[p].val_v_av);
        p=cen[p].par;
    }
}

void dell(LL x) {
    LL p=x;
    while (p!=-1) {
        cen[p].val_v -= {x,0};
        cen[p].val_v_av -= {x*dis[cen[p].depth][x],0};
        if (cen[p].par != -1) {
            int par=cen[p].par;
            cen[p].val_v += {0,x};
            cen[p].val_v_av += {0,x*dis[cen[par].depth][x]};
        }
        Pure(cen[p].val_v);
        Pure(cen[p].val_v_av);
        p=cen[p].par;
    }
}

LL query(LL x) {
    LL ret=0;
    LL v=0;
    LL v_av=0;
    int p=x;
    while (p!=-1) {
        v += cen[p].val_v.first;
        v_av += cen[p].val_v_av.first;
        ret += x*v_av;
        ret %= mod;
        ret += x*dis[cen[p].depth][x]*v;
        ret %= mod;
        v = cen[p].val_v.second;
        v_av = cen[p].val_v_av.second;
        p=cen[p].par;
    }
    return ret;
}

LL pow(LL a,LL n,LL mod) {
    if (n==0) return 1;
    else if (n==1) return a;
    LL ret=pow(a,n/2,mod);
    ret*=ret;
    ret%=mod;
    if (n&1) {
        ret*=a;
        ret%=mod;
    }
    return ret;
}

int main () {
    int n,q;
    scanf("%d %d",&n,&q);
    for (int i=1;n-1>=i;i++) {
        int a,b;
        scanf("%d %d",&a,&b);
        edg[a].push_back(b);
        edg[b].push_back(a);
    }
    dfs(1,-1,0);
    while (q--) {
        int k;
        scanf("%d",&k);
        vector<int> v;
        while (k--) {
            int x;
            scanf("%d",&x);
            v.push_back(x);
        }
        for (int i:v) addd(i);
        LL ans=0;
        for (int i:v) {
            ans += query(i);
            ans%=mod;
        }
        for (int i:v) dell(i);
        printf("%lld\n",(ans*pow(2,mod-2,mod) + mod)%mod);
    }
}