Submission #917157

#	Time	Username	Problem	Language	Result	Execution time	Memory
917157		GrindMachine	Cats or Dogs (JOI18_catdog)	C++17	100 / 100	563 ms	22352 KiB

catdog

#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;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x,y) ((x+y-1)/(y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep1(i,n) for(int i = 1; i <= n; ++i)
#define rev(i,s,e) for(int i = s; i >= e; --i)
#define trav(i,a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a,b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a,b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif

/*

refs:
edi
https://oj.uz/submission/374896

tree dp for 38 points:
r[u] = sum(min(r[v],b[v]+1))
b[u] = sum(min(r[v]+1,b[v]))

key idea:
when a node is changed, only the dp values of its parents are affected
only update parents
=> hld

if the problem was on a line, we could use a segtree
each node [l,r] contains dp[x][y], which denotes the min cost to achieve comp[l] = x and comp[r] = y
can be merged easily

how to extend this idea to a tree?

each node belongs to exactly 1 chain in the hld
when processing a chain, only dp values in the chain will change
some nodes on the chain may have children that belong to other chains
for such children, their values wont change, so their contribution to dp[u][0/1] is fixed
because these values dont change, we can put them in the segtree leaf that denotes u
i.e for the segtree leaf that denotes u,
dp[0][0] = sum(min(r[v],b[v]+1)), v doesnt belong to the same chain as u
dp[1][1] = sum(min(r[v]+1,b[v])), v doesnt belong to the same chain as u
dp[0][1] = dp[1][0] = inf (range only contains 1 node, so starting comp = ending comp)

with all these values in the segtree, find the value at the root of the chain
now when we move up, we move to another chain
so we have to update the new dp values of the parent of the current chain (which may or may not be the root of the new chain)
tnis can be done by adding/subtracting some value from the dp values of the parent and then doing a point update on the segtree
repeat the same process for all chains
at the end of the process, we would have updated the dp chain that contains the root of the tree (which is 1)
when we want to get the answer, just find the dp of the chain that contains the root and return the min value of dp[x][y]

*/

const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

template<typename T>
struct segtree {
    // https://codeforces.com/blog/entry/18051

    /*=======================================================*/

    struct data {
        array<array<int,2>,2> dp;
        bool active;

        data(){
            rep(i,2){
                rep(j,2){
                    dp[i][j] = inf1;
                }
            }
            active = false;
        }
    };

    data neutral = data();

    data merge(data &left, data &right) {
        if(!left.active and !right.active) return left;
        if(!right.active) return left;
        if(!left.active) return right;
        
        data curr;
        curr.active = true;

        rep(i,2){
            rep(j,2){
                rep(k,2){
                    rep(l,2){
                        amin(curr.dp[i][l],left.dp[i][j]+right.dp[k][l]+(j!=k));
                    }
                }
            }
        }

        return curr;
    }

    void create(int i, T v) {

    }

    void modify(int i, T v) {
        tr[i] = neutral;
        tr[i].dp[0][0] = v.ff;
        tr[i].dp[1][1] = v.ss;
        tr[i].active = true;
    }

    /*=======================================================*/

    int n;
    vector<data> tr;

    segtree() {

    }

    segtree(int siz) {
        init(siz);
    }

    void init(int siz) {
        n = siz;
        tr.assign(2 * n, neutral);
    }

    void build(vector<T> &a, int siz) {
        rep(i, siz) create(i + n, a[i]);
        rev(i, n - 1, 1) tr[i] = merge(tr[i << 1], tr[i << 1 | 1]);
    }

    void pupd(int i, T v) {
        modify(i + n, v);
        for (i = (i + n) >> 1; i; i >>= 1) tr[i] = merge(tr[i << 1], tr[i << 1 | 1]);
    }

    data query(int l, int r) {
        data resl = neutral, resr = neutral;

        for (l += n, r += n; l <= r; l >>= 1, r >>= 1) {
            if (l & 1) resl = merge(resl, tr[l++]);
            if (!(r & 1)) resr = merge(tr[r--], resr);
        }

        return merge(resl, resr);
    }
};

vector<int> adj[N];
vector<int> a(N); // 0 = none, 1 = cat, 2 = dog
vector<int> subsiz(N);
vector<int> depth(N), par(N);

void dfs1(int u, int p){
    subsiz[u] = 1;
    if(p != -1) par[u] = p;
    trav(v,adj[u]){
        if(v == p) conts;
        depth[v] = depth[u]+1;
        dfs1(v,u);
        subsiz[u] += subsiz[v];
    }
}

vector<int> pos(N), head(N), chain_siz(N);
int timer = 1;

void dfs2(int u, int p, int h){
    pos[u] = timer++;
    head[u] = h;
    chain_siz[h]++;

    pii mx = {-inf1,-1};
    trav(v,adj[u]){
        if(v == p) conts;
        pii px = {subsiz[v],v};
        amax(mx,px);
    }

    int heavy = mx.ss;
    if(heavy != -1){
        dfs2(heavy,u,h);
    }

    trav(v,adj[u]){
        if(v == p or v == heavy) conts;
        dfs2(v,u,v);
    }
}

segtree<pii> st;

void initialize(int n, std::vector<int> A, std::vector<int> B) {
    rep(i,n-1){
        int u = A[i], v = B[i];
        adj[u].pb(v), adj[v].pb(u);
    }

    dfs1(1,-1);
    dfs2(1,-1,1);
    st = segtree<pii>(n+5);
    rep1(i,n) st.pupd(i,{0,0});
}

vector<int> sum1(N), sum2(N);

int get_ans(){
    auto dp = st.query(pos[1],pos[1]+chain_siz[1]-1).dp;

    int ans = inf1;

    rep(i,2){
        rep(j,2){
            amin(ans,dp[i][j]);
        }
    }

    return ans;
}

void rem(int u){
    while(u){
        if(u == head[u]){
            auto dp = st.query(pos[u],pos[u]+chain_siz[u]-1).dp;
            int cat = min(dp[0][0],dp[0][1]);
            int dog = min(dp[1][0],dp[1][1]);
            sum1[par[u]] -= min(cat,dog+1);
            sum2[par[u]] -= min(cat+1,dog);
            u = par[u];
        }
        else{
            u = head[u];
        }
    }
}

void add(int u){
    while(u){
        {
            pii px = {sum1[u],sum2[u]};

            if(a[u] == 1){
                px.ss = inf1;
            }
            else if(a[u] == 2){
                px.ff = inf1;
            }

            st.pupd(pos[u],px);
        }

        if(u == head[u]){
            auto dp = st.query(pos[u],pos[u]+chain_siz[u]-1).dp;
            int cat = min(dp[0][0],dp[0][1]);
            int dog = min(dp[1][0],dp[1][1]);
            sum1[par[u]] += min(cat,dog+1);
            sum2[par[u]] += min(cat+1,dog);
            u = par[u];
        }
        else{
            u = head[u];
        }
    }
}

void change_state(int u, int val){
    rem(u);
    a[u] = val;
    add(u);
}
 
int cat(int v) {
    change_state(v,1);
    return get_ans();
}
 
int dog(int v) {
    change_state(v,2);
    return get_ans();
}
 
int neighbor(int v) {
    change_state(v,0);
    return get_ans();
}

• Subtask 1: 8 / 8
• Subtask 2: 30 / 30
• Subtask 3: 62 / 62