Submission #1119948

#	Time	Username	Problem	Language	Result	Execution time	Memory
1119948		luvna	Election Campaign (JOI15_election_campaign)	C++17	30 / 100	1071 ms	64856 KiB

election_campaign

#include<bits/stdc++.h>

#define MASK(i) (1 << (i))
#define pub push_back
#define all(v) v.begin(), v.end()
#define compact(v) v.erase(unique(all(v)), end(v))
#define pii pair<int,int>
#define fi first
#define se second
#define endl "\n"
#define dbg(x) "[" #x " = " << (x) << "]"

using namespace std;

template<class T> bool minimize(T& a, T b){if(a > b) return a = b, true;return false;}
template<class T> bool maximize(T& a, T b){if(a < b) return a = b, true;return false;}

typedef long long ll;
typedef long double ld;

mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
ll rand(ll l, ll r){return uniform_int_distribution<ll>(l, r)(rng);}

const int N = 2e5 + 15;
const int lg = 16;

struct node{
    int u, v, w;
};

struct FenwickTree{
    int n;
    vector<ll> bit;

    FenwickTree(int n) : n(n), bit(n + 15, 0) {}

    void update(int idx, ll v){
        for(;idx <= n; idx += (idx & (-idx))) bit[idx] += v;
    }

    ll get(int l, int r){ll res = 0; l--;
        for(;r;r -= (r & (-r))) res += bit[r];
        for(;l;l -= (l & (-l))) res -= bit[l];
        return res;
    }
};

vector<int> g[N];
int head[N], hev[N], par[N][lg+1];
int tin[N], h[N], sz[N], timerDFS;
vector<pii> evtOne[N];
vector<node> evtTwo[N];
ll dp[N], sum[N];

void solve(){
    int n; cin >> n;

    for(int i = 1; i < n; i++){
        int u, v; cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }

//    vector<int> head(n+15, 0), hev(n + 15, 0);
//    vector<vector<int>> par(n + 15, vector<int>(lg+2));
//    vector<int> tin(n + 15, 0), sz(n + 15, 0), h(n + 15, 0);
    auto dfs = [&](auto& dfs, int u) -> void{
        sz[u] = 1;
        for(int v : g[u]) if(v != par[u][0]){
            h[v] = h[u] + 1;
            par[v][0] = u;
            for(int j = 1; j <= lg; j++) par[v][j] = par[par[v][j-1]][j-1];
            dfs(dfs, v);
            if(sz[v] > sz[hev[u]]) hev[u] = v;
        }
    };

    auto hld = [&](auto& hld, int u, int cur_chain) -> void{
        head[u] = cur_chain;
        tin[u] = ++timerDFS;
        if(hev[u]) hld(hld, hev[u], cur_chain);
        for(int v : g[u]) if(v != par[u][0] && v != hev[u]) hld(hld, v, v);
    };

    dfs(dfs,1);
    hld(hld, 1, 1);

    FenwickTree bit(n);

    auto lca = [&](int u, int v) -> int{
        if(h[u] < h[v]) swap(u,v);

        for(int j = lg; j >= 0; j--) if(h[u] - (1 << j) >= h[v]){
            u = par[u][j];
        }

        if(u == v) return u;

        for(int j = lg; j >= 0; j--){
            if(par[u][j] != par[v][j]){
                u = par[u][j];
                v = par[v][j];
            }
        }

        return par[u][0];
    };

    //2 case:
    //lca(u,v) = u -> query on 1 subtree
    //lca(u,v) != u and v -> query on 2 subtree

    int m; cin >> m;

//    vector<vector<pii>> evtOne(n + 15);
//    vector<vector<node>> evtTwo(n + 15);

    while(m--){
        int u, v, w; cin >> u >> v >> w;
        if(h[u] > h[v]) swap(u,v);
        int LCA = lca(u,v);
        //cout << u << " " << v << "  " << lca(u,v) << endl;
        if(u == LCA) evtOne[LCA].push_back({v,w});
        else evtTwo[LCA].push_back({u,v,w});
    }

    auto get = [&](int u, int v) -> ll{
        ll res = 0;
        for(;head[u] != head[v]; u = par[head[u]][0]){
            if(h[head[u]] < h[head[v]]) swap(u,v);
            res += bit.get(tin[head[u]], tin[u]);
        }
        if(h[u] > h[v]) swap(u,v);
        res += bit.get(tin[u], tin[v]);
        return res;
    };

    auto bin_lift = [&](int u, int v) -> int{
        for(int j = lg; j >= 0; j--){
            if(h[v] - (1 << j) > h[u]) v = par[v][j];
        }
        return v;
    };

    auto dfsDP = [&](auto& dfsDP, int u) -> void{
        for(int v : g[u]) if(v != par[u][0]){
            dfsDP(dfsDP, v);
            sum[u] += dp[v];
        }

        //cout << "case: " << dbg(u) << endl << endl;

        dp[u] = sum[u];

        //case 1: u = lca(u,v): sum[u] + sum[v] + w + cost(u -> v)
        for(auto &[v, w] : evtOne[u]){
            ll cost = sum[u] + w;
            cost -= dp[bin_lift(u,v)];
            //cout << v << " " << bin_lift(u,v) << endl;
            //cout << u << " " << v << " bel " << get(u,v) << endl;
            cost += get(u,v);
            cost += sum[v];
            maximize(dp[u], cost);
        }

        //case 2:
        //add cost(u -> U) + cost(u -> V)
        //subtract dp[bin_lift(U and V)]
        // cost(u -> U) sum[u] - dp[v]: get down to v without adding dp[v]
        for(auto &[U, V, w] : evtTwo[u]){
            ll cost = sum[u] + w;
            cost -= dp[bin_lift(u,U)];
            cost -= dp[bin_lift(u,V)];
            cost += sum[U];
            cost += sum[V];
            cost += get(u,U);
            cost += get(u,V);
            maximize(dp[u], cost);
        }

        //cout << dbg(u) << dbg(dp[u]) << dbg(sum[u]) << endl;

        for(int v : g[u]) if(v != par[u][0]){
            bit.update(tin[v], sum[u] - dp[v]);
        }
    };

    dfsDP(dfsDP, 1);

    cout << dp[1];
}

signed main(){
    ios_base::sync_with_stdio(NULL);
    cin.tie(0); cout.tie(0);

    #define task "task"

    if(fopen(task".INP", "r")){
        freopen(task".INP", "r", stdin);
        freopen(task".OUT", "w", stdout);
    }

    int t; t = 1; //cin >> t;
    while(t--) solve();
}

Compilation message (stderr)

election_campaign.cpp: In function 'int main()':
election_campaign.cpp:200:16: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
  200 |         freopen(task".INP", "r", stdin);
      |         ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~
election_campaign.cpp:201:16: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
  201 |         freopen(task".OUT", "w", stdout);
      |         ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~

• Subtask 1: 10 / 10
• Subtask 2: 5 / 5
• Subtask 3: 5 / 5
• Subtask 4: 0 / 30
• Subtask 5: 10 / 10
• Subtask 6: 0 / 40