2025年11月22日训练赛

F1. Cycling (Easy Version)

#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#define yes cout << "Yes" << endl
#define no cout << "No" << endl
#define pii pair<int,int>
#define ll long long
#define pb push_back
#define ft first
#define se second
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f
#define int long long

const int N=5010;
int n;int a[N];int f[N];
void solve(){
cin>>n;
for(int i=1;i<=n;i++)cin>>a[i];

for(int i=1;i<=n;i++)f[i]=INF;
f[0]=0;
for(int i=1;i<=n;i++){
    int mx=i;
    for(int j=i;j>=1;j--){
        if(a[j]<a[mx])mx=j;
        f[i]=min(f[i],f[j-1]+i-mx+i-j+a[mx]*(i-j+1));
    }
}
cout<<f[n]<<'\n';

}
signed main(){
    std::ios::sync_with_stdio(false);
    int T=1;cin>>T;
    while(T--){
        solve();
    }
}

B. Gellyfish and Baby's Breath

签到

cf786B B. Legacy

线段树优化建图板子
线段树的结点作为图的点，建树，出树儿子向父亲，入树父亲向儿子。两树的叶子结点相连

#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#define yes cout << "Yes" << endl
#define no cout << "No" << endl
#define pii pair<int,int>
#define ll long long
#define pb push_back
#define ft first
#define se second
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f

#define int long long

const int N = 100010;
int n, q, s; int lf[N];
vector<pii> G[2 * 4 *N];
int d[2 * 4 * N];
struct Tree{
    int l,r;
}tr[4*N];
void bui(int u, int l, int r){
    tr[u] = {l, r};
    if(l == r){
        lf[l] = u;//存叶子
        return ;
    }

    //出树，儿子向父亲
    G[u * 2 + 4 * n].pb({u + 4 * n, 0});
    G[u * 2 + 1 + 4 * n].pb({u + 4 * n,0});
    //入树，父亲向儿子
    G[u].pb({u * 2, 0});
    G[u].pb({u * 2 + 1, 0});

    int mid = (l + r) >> 1;
    bui(u*2, l, mid);
    bui(u*2 + 1, mid + 1, r);
}

void query(int u, int x, int y, int v, int w, int rev){
    if(x <= tr[u].l && tr[u].r <= y){
        //出树向入树
        if(rev == 0) G[v + 4*n].pb({u, w});
        else G[u + 4*n].pb({v,w});
        return ;
    }
    int mid = (tr[u].l + tr[u].r) >> 1;
    if(x <= mid) query(u << 1, x, y, v, w, rev);
    if(y >= mid + 1)query(u << 1 | 1, x, y, v, w, rev);
}

void dijk(){
    priority_queue<pii, vector<pii>, greater<pii> > pq;
    for(int i = 1; i <= 4*n + 4*n; i ++) d[i] = INF;
    d[lf[s]] = 0; pq.push({0, lf[s]});

    while(!pq.empty()){
        auto [_, u] = pq.top(); pq.pop();

        for(auto [v,w] : G[u]){
            if(d[v] > d[u] + w){
                d[v] = d[u] + w;
                pq.push({d[v], v});
            }
        }
    }

}

void solve(){
     cin >> n >> q >> s;

    bui(1, 1, n);
    for(int i = 1; i <= n; i ++){
        G[lf[i]].pb({lf[i] + 4*n, 0});
        G[lf[i] + 4*n].pb({lf[i], 0});
    }

    while(q --){
        int op; cin >> op;
        if(op == 1){
            int v, u, w; cin >> v >> u >> w;
            G[lf[v]].pb({lf[u], w});
        }else if(op == 2){
            int v, l, r, w; cin >> v >> l >> r >> w;
            query(1, l, r, lf[v], w, 0); 
        }else {
            int v, l, r, w; cin >> v >> l >> r >> w;
            query(1, l, r, lf[v], w, 1);
        }
    }
    dijk();

    for(int i = 1; i <= n; i ++){
        if(d[lf[i]] > INF / 2) cout << -1 << ' ';
        else cout << d[lf[i]] << ' ';
    } cout << '\n';

}
signed main(){
    std::ios::sync_with_stdio(false);
    int T = 1; // cin >> T;
    while(T--){
        solve();
    }
}