Educational Codeforces Round 36

Educational Codeforces Round 36 

D. Almost Acyclic Graph

首先想到枚举所有的边,然后判断删掉之后是否还有环,这样的话复杂度会到\(O(m^2)\),发现\(n\)很小,那么最大的环大小不会超过\(n\),所以考虑找到一个环,然后枚举环上的边删掉之后判断是否还有环

判环可以用拓扑排序来做

view code
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast,no-stack-protector")
#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define endl "\n"
#define LL long long int
#define vi vector<int>
#define vl vector<LL>
#define all(V) V.begin(),V.end()
#define sci(x) scanf("%d",&x)
#define scl(x) scanf("%I64d",&x)
#define scs(s) scanf("%s",s)
#define pii pair<int,int>
#define pll pair<LL,LL>
#ifndef ONLINE_JUDGE
#define cout cerr
#endif
#define cmax(a,b) ((a) = (a) > (b) ? (a) : (b))
#define cmin(a,b) ((a) = (a) < (b) ? (a) : (b))
#define debug(x)  cerr << #x << " = " << x << endl
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
template <typename T> vector<T>& operator << (vector<T> &__container, T x){ __container.push_back(x); return __container; }
template <typename T> ostream& operator << (ostream &out, vector<T> &__container){ for(T _ : __container) out << _ << ' '; return out; }
const int MAXN = 5e2+7;
vi G[MAXN];
bool vis[MAXN], instk[MAXN];
stack<int> stk;
vector<int> vec;
void dfs(int u){
    vis[u] = true;
    stk.push(u);
    instk[u] = true;
    for(int v : G[u]){
        if(!vis[v]) dfs(v);
        else if(instk[v]){
            do{
                vec << stk.top();
                stk.pop();
            }while(vec.back()!=v);
            reverse(all(vec));
        }
        if(!vec.empty()) return;
    }
    instk[u] = false;
    stk.pop();
}
void solve(){
    int n, m;
    sci(n); sci(m);
    for(int i = 1; i <= m; i++){
        int u, v; sci(u); sci(v);
        G[u] << v;
    }
    for(int i = 1; i <= n; i++) if(!vis[i]){
        dfs(i);
        if(!vec.empty()) break;
    }
    if(vec.empty()){
        cout << "YES" << endl;
        return;
    }
    bool ok = false;
    auto test = [&](int a, int b){
        vi deg(n+1,0);
        for(int i = 1; i <= n; i++) for(int v : G[i]){
            if(i==a and v==b) continue;
            deg[v]++;
        }
        int used = 0;
        queue<int> que;
        for(int i = 1; i <= n; i++) if(!deg[i]) que.push(i);
        while(!que.empty()){
            int u = que.front();
            que.pop(), used++;
            for(int v : G[u]){
                if(u==a and v==b) continue;
                if(!--deg[v]) que.push(v);
            }
        }
        return used==n;
    };
    for(int i = 0; i < vec.size() - 1; i++){
        if(test(vec[i],vec[i+1])){
            ok = true;
            break;
        }
    }
    if(test(vec.back(),vec.front())) ok = true;
    cout << (ok ? "YES" : "NO") << endl;
}
int main(){
    #ifndef ONLINE_JUDGE
    freopen("Local.in","r",stdin);
    freopen("ans.out","w",stdout);
    #endif
    solve();
    return 0;
}

E. Physical Education Lessons

那么对于这种区间范围很大的数据结构题,一般离散化之后用线段树来做,或者动态开点线段树

这里考虑离散化之后用线段树来做,注意把每个要离散化点的右端点也加进去,保证这个点在线段树上的区间范围是\(1\)

view code
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast,no-stack-protector")
#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define endl "\n"
#define LL long long int
#define vi vector<int>
#define vl vector<LL>
#define all(V) V.begin(),V.end()
#define sci(x) scanf("%d",&x)
#define scl(x) scanf("%I64d",&x)
#define scs(s) scanf("%s",s)
#define pii pair<int,int>
#define pll pair<LL,LL>
#ifndef ONLINE_JUDGE
#define cout cerr
#endif
#define cmax(a,b) ((a) = (a) > (b) ? (a) : (b))
#define cmin(a,b) ((a) = (a) < (b) ? (a) : (b))
#define debug(x)  cerr << #x << " = " << x << endl
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
template <typename T> vector<T>& operator << (vector<T> &__container, T x){ __container.push_back(x); return __container; }
template <typename T> ostream& operator << (ostream &out, vector<T> &__container){ for(T _ : __container) out << _ << ' '; return out; }
const int MAXN = 1.2e6+7;
int n, q;
vector<int> vec;
struct STA{
    int l, r, k;
}sta[MAXN];
struct SegmentTree{
    int l[MAXN<<2], r[MAXN<<2], sum[MAXN<<2], lazy[MAXN<<2];
    #define ls(rt) rt << 1
    #define rs(rt) rt << 1 | 1
    #define pushup(rt) sum[rt] = sum[ls(rt)] + sum[rs(rt)]
    void build(int L, int R, int rt = 1){
        l[rt] = L; r[rt] = R;
        lazy[rt] = -1;
        if(L + 1 == R){
            sum[rt] = vec[r[rt]] - vec[l[rt]];
            return;
        }
        int mid = (L + R) >> 1;
        build(L,mid,ls(rt)); build(mid,R,rs(rt));
        pushup(rt);
    }
    void pushdown(int rt){
        if(lazy[rt]==-1) return;
        lazy[ls(rt)] = lazy[rs(rt)] = lazy[rt];
        if(lazy[rt]==1){
            sum[ls(rt)] = vec[r[ls(rt)]] - vec[l[ls(rt)]];
            sum[rs(rt)] = vec[r[rs(rt)]] - vec[l[rs(rt)]];
        }else sum[ls(rt)] = sum[rs(rt)] = 0;
        lazy[rt] = -1;
        return;
    }
    void modify(int L, int R, int x, int rt = 1){
        if(L>=r[rt] or l[rt]>=R) return;
        if(L<=l[rt] and r[rt]<=R){
            lazy[rt] = x;
            if(x==1) sum[rt] = vec[r[rt]] - vec[l[rt]];
            else sum[rt] = 0;
            return;
        }
        pushdown(rt);
        modify(L,R,x,ls(rt)); modify(L,R,x,rs(rt));
        pushup(rt);
    }
    int query(int L, int R, int rt = 1){
        if(L>=r[rt] or l[rt]>=R) return 0;
        if(L<=l[rt] and r[rt]<=R) return sum[rt];
        pushdown(rt);
        return query(L,R,ls(rt)) + query(L,R,rs(rt));
    }
}ST;
void solve(){
    sci(n); sci(q);
    for(int i = 1; i <= q; i++){
        sci(sta[i].l); sci(sta[i].r); sci(sta[i].k);
        vec << sta[i].l << sta[i].r;
        if(sta[i].l+1<=n) vec << sta[i].l+1;
        if(sta[i].r+1<=n) vec << sta[i].r+1;
    }
    sort(all(vec)); vec.erase(unique(all(vec)),vec.end());
    vec.push_back(vec.back()+1);
    for(int i = 1; i <= q; i++){
        sta[i].l = lower_bound(all(vec),sta[i].l) - vec.begin();
        sta[i].r = lower_bound(all(vec),sta[i].r) - vec.begin();        
    }
    int ret = n;
    ST.build(0,vec.size()-1);
    for(int i = 1; i <= q; i++){
        int k = sta[i].k, l = sta[i].l, r = sta[i].r;
        ret -= ST.query(l,r+1); 
        ST.modify(l,r+1,k-1);
        if(k==2) ret += vec[r] - vec[l] + 1;
        cout << ret << endl;
    }
}
int main(){
    #ifndef ONLINE_JUDGE
    freopen("Local.in","r",stdin);
    freopen("ans.out","w",stdout);
    #endif
    solve();
    return 0;
}

F. Imbalance Value of a Tree

首先,可以把问题转化为任意两点间最大值的和减去最小值的和

以最大值为例,最小值类似

考虑类似点分治的思想,每次找到所有点中的最大值,然后计算这个点的贡献就是经过这个点的所有路径的数量乘上点权,然后把这个点删掉之后变成几个子树,分治解决,但是每次找到的点不是树的重心,所以分治复杂度最坏会到达\(O(n^2)\)

考虑反着来做,每次找到当前最小的点,然后把连着的所有比他小的点都连起来,然后算当前连通块的贡献,可以用并查集来做,复杂度为\(O(n\log n)\)

对于某个连通块,假设当前的最大值点为\(u\),那么路径的总数应该等于任意两个子树的大小的乘积加上总的大小,前者可以用前缀和来优化计算

view code
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast,no-stack-protector")
#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define endl "\n"
#define LL long long int
#define vi vector<int>
#define vl vector<LL>
#define all(V) V.begin(),V.end()
#define sci(x) scanf("%d",&x)
#define scl(x) scanf("%I64d",&x)
#define scs(s) scanf("%s",s)
#define pii pair<int,int>
#define pll pair<LL,LL>
#ifndef ONLINE_JUDGE
#define cout cerr
#endif
#define cmax(a,b) ((a) = (a) > (b) ? (a) : (b))
#define cmin(a,b) ((a) = (a) < (b) ? (a) : (b))
#define debug(x)  cerr << #x << " = " << x << endl
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
template <typename T> vector<T>& operator << (vector<T> &__container, T x){ __container.push_back(x); return __container; }
template <typename T> ostream& operator << (ostream &out, vector<T> &__container){ for(T _ : __container) out << _ << ' '; return out; }
const int MAXN = 1e6+7;
int n, app[MAXN], root[MAXN], sz[MAXN];
pair<pii,int> w[MAXN], cp[MAXN];
vi G[MAXN];
int findx(int x){ return x==root[x]?x:root[x]=findx(root[x]); }


void solve(){
    sci(n);
    for(int i = 1; i <= n; i++){
        sci(w[i].first.first);
        w[i].first.second = ++app[w[i].first.first];
        w[i].second = i;
        cp[i] = w[i];
    }
    for(int i = 1; i < n; i++){
        int u, v;
        sci(u); sci(v);
        G[u] << v; G[v] << u;
    }
    sort(cp+1,cp+1+n);
    LL ret = 0;
    for(int i = 1; i <= n; i++) sz[i] = 1, root[i] = i;
    for(int i = 1; i <= n; i++){
        vi vec, sum;
        int u = cp[i].second;
        for(int v : G[u]){
            if(w[v].first < w[u].first){
                vec << sz[findx(v)];
                sz[findx(u)] += sz[findx(v)];
                root[findx(v)] = findx(u);
            }
        }
        partial_sum(all(vec),back_inserter(sum));
        LL tot = 0;
        for(int j = 1; j < vec.size(); j++) tot += 1ll * vec[j] * sum[j-1];
        tot += sz[findx(u)];
        ret += w[u].first.first * tot;
    }
    for(int i = 1; i <= n; i++) sz[i] = 1, root[i] = i;
    for(int i = n; i >= 1; i--){
        vi vec, sum;
        int u = cp[i].second;
        for(int v : G[u]){
            if(w[v].first > w[u].first){
                vec << sz[findx(v)];
                sz[findx(u)] += sz[findx(v)];
                root[findx(v)] = findx(u);
            }
        }
        partial_sum(all(vec),back_inserter(sum));
        LL tot = 0;
        for(int j = 1; j < vec.size(); j++) tot += 1ll * vec[j] * sum[j-1];
        tot += sz[findx(u)];
        ret -= w[u].first.first * tot;
    }
    cout << ret << endl;
}
int main(){
    #ifndef ONLINE_JUDGE
    freopen("Local.in","r",stdin);
    freopen("ans.out","w",stdout);
    #endif
    solve();
    return 0;
}

G. Coprime Arrays

要计算\(\sum_{i=1}^k (\sum_{a_1=1}^i\cdots \sum_{a_n=1}^i[\operatorname {gcd}(a_1\cdots a_n)=1])\oplus i\)

考虑单独计算\(g(i) = \sum_{a_1=1}^i\cdots \sum_{a_n=1}^i[\operatorname {gcd}(a_1\cdots a_n)=1]\)

通过莫比乌斯反演可以得到:\(g(i) = \sum_{a_1=1}^i\cdots \sum_{a_n=1}^i[\operatorname {gcd}(a_1\cdots a_n)=1] = \sum_{j=1}^i\mu(j)\cdot \lfloor \frac ij\rfloor^n\)

如果直接用整除分块来做的话,复杂度是\(O(\sum_{i=1}^k\sqrt i) = O(\frac 23 k^{\frac 32})\)

考虑相邻两个\(g(i),g(i-1)\)可以发现只有满足\(j\mid i\)\(\lfloor \frac ij\rfloor\)发生了变化,所以考虑计算\(f(i) = g(i)-g(i-1)\),所有数的因子数的复杂度为\(O(k\log k)\),再预处理一下所有的幂,总复杂度为\(O(k(\log n + \log k))\)

view code
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast,no-stack-protector")
#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define endl "\n"
#define LL long long int
#define vi vector<int>
#define vl vector<LL>
#define all(V) V.begin(),V.end()
#define sci(x) scanf("%d",&x)
#define scl(x) scanf("%I64d",&x)
#define scs(s) scanf("%s",s)
#define pii pair<int,int>
#define pll pair<LL,LL>
#ifndef ONLINE_JUDGE
#define cout cerr
#endif
#define cmax(a,b) ((a) = (a) > (b) ? (a) : (b))
#define cmin(a,b) ((a) = (a) < (b) ? (a) : (b))
#define debug(x)  cerr << #x << " = " << x << endl
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
template <typename T> vector<T>& operator << (vector<T> &__container, T x){ __container.push_back(x); return __container; }
template <typename T> ostream& operator << (ostream &out, vector<T> &__container){ for(T _ : __container) out << _ << ' '; return out; }
const int MAXN = 2e6+7;
const int MOD = 1e9+7;
int mu[MAXN], prime[MAXN], pri_cnt, pw[MAXN];
int f[MAXN];
bool npm[MAXN];
LL ksm(LL a, LL b){
    LL ret = 1;
    while(b){
        if(b&1) ret = ret * a % MOD;
        b >>= 1;
        a = a * a % MOD;
    }
    return ret;
}
void sieve(){
    mu[1] = 1;
    for(int i = 2; i < MAXN; i++){
        if(!npm[i]) prime[++pri_cnt] = i, mu[i] = -1;
        for(int j = 1; i * prime[j] < MAXN; j++){
            npm[i*prime[j]] = true;
            if(i%prime[j]==0){
                mu[i*prime[j]] = 0;
                break;
            }
            mu[i*prime[j]] = -mu[i];
        }
    }
}
void solve(){
    sieve();
    int n, k;
    sci(n); sci(k);
    for(int i = 1; i <= k; i++) pw[i] = ksm(i,n);
    for(int i = 1; i <= k; i++) for(int j = i; j <= k; j += i) f[j] = (f[j] + mu[i] * (pw[j/i] - pw[j/i-1])) % MOD;
    int ret = 0, cur = 0;
    for(int i = 1; i <= k; i++){
        cur = ((cur + f[i]) % MOD + MOD) % MOD;
        ret = (ret + (cur ^ i)) % MOD;
    }
    cout << ret << endl;
}
int main(){
    #ifndef ONLINE_JUDGE
    freopen("Local.in","r",stdin);
    freopen("ans.out","w",stdout);
    #endif
    solve();
    return 0;
}
posted @ 2020-08-31 17:16  _kiko  阅读(42)  评论(0编辑  收藏