[CSP-S 2023] 种树

 

 

 

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#include<bits/stdc++.h> #define ll long long #define pb push_back #define mxn 100003 #define rep(i,a,b) for(int i=a;i<=b;++i) #define rept(i,a,b) for(int i=a;i<b;++i) using namespace std; int n,p[mxn],d[mxn],ct[mxn]; ll a[mxn],b[mxn],c[mxn]; vector<int>g[mxn]; bool v[mxn]; inline __int128 get(ll i,ll n,__int128 a,__int128 b) //从第i天到第N天，按试题的规定，某棵树能长多高 // b[i]+x*C[i],其中a代表c[i],b代表b[i]   {     if(a<0)     {         ll d=min((b-a-1)/(-a),(__int128)n+1);         if(d<=i)            return n-i+1;         return n-d+1+(d-i)*b+(d-1+i)*(d-i)/2*a;     }     return (n+i)*(n-i+1)/2*a+b*(n-i+1); } void dfs(int x,int fa) {     for(int i:g[x])        if(i!=fa)        {                 dfs(i,x);                 p[x]=min(p[x],p[i]-1);                 //对于第x个树来说，p[x]代表它在第几天就应该种上来                 //但由于只有种下x，才能种它的子结点，于是取它自己的时间与子结点时间-1的较小值        }  } bool check(int mx) {     rep(i,1,n)     //算出每棵树，最晚应该在哪个时间点来种     {         int l=1,r=n;         //二分找第i棵树最迟种下来的时间，R的初值必须为n         //如果设为mx的话，会过界的         int ans=0;         while(l<=r)         {             int mid=(l+r)>>1;             if(a[i]<=get(mid,mx,c[i],b[i]))                 ans=max(ans,mid),l=mid+1;             else                 r=mid-1;         }         if (ans==0) return 0;         if(i==1)            ans=1;         p[i]=ans;     }   //  for(int i=1;i<=n;i++)     //     cout<<i<<"   "<<p[i]<<endl;     dfs(1,0);      //修正每个点最晚的种树时间，修正后p数组的值会控制在1到N之间           rep(i,1,n)        ct[i]=0;     rep(i,1,n)     {         if(p[i]<1) //这句必须要加上，因为有可能修正后p[1]变成了0             return 0;         ct[p[i]]++; //在第p[i]个时间要种树的行为，要执行多少次     }     rep(i,1,n)     //枚举时间     {         ct[i]+=ct[i-1];  //统计一共要种多少棵树         if(ct[i]>i) //前i个时间只能种i棵树            return 0;     }     return 1; } signed main(){     scanf("%d",&n);     rep(i,1,n)         scanf("%lld%lld%lld",&a[i],&b[i],&c[i]);     for(int i=1,x,y;i<n;++i)     {         scanf("%d%d",&x,&y);         g[x].pb(y),g[y].pb(x);     }     int l=n,r=1e9;     int ans=1e9;     while(l<=r)     {         int mid=(l+r)>>1;         if(check(mid))            ans=min(ans,mid),r=mid-1;         else l=mid+1;     }     cout<<ans<<endl;     return 0; }

　　

 

　　

 

 

 

 

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#include <cstdio> #include <iostream> #include <algorithm> #include <vector> #include <map> #include <queue> using namespace std;   const int N = 1e5, E = N << 1; const long long Max = 1e9;   typedef pair<long long, int> pir;   int n; long long a[N + 5], b[N + 5], c[N + 5], zero[N + 5];   int head[N + 5], to[E + 5], nxt[E + 5], tot = 1; void add_edge(int u, int v) {     tot++;     to[tot] = v;     nxt[tot] = head[u];     head[u] = tot;     return ; } void add(int u, int v) {     add_edge(u, v);     add_edge(v, u);     return ; }   long long d[N + 5];//limit int sz[N + 5];   void calc_d(int u, long long ans) {     long long l = 1, r = n;     d[u] = -1ll;     while (l <= r)     {         long long mid = (l + r) >> 1;         __int128 sum = 0, one = 1;           if (c[u] >= 0)             sum = one * (ans - mid + 1) * b[u]                  + one * (mid + ans) * (ans - mid + 1) / 2 * c[u];         else         {             if (mid > zero[u])                 sum = ans - mid + 1;             else if (ans > zero[u])                 sum = one * (zero[u] - mid + 1) * b[u]                      + one * (mid + zero[u]) * (zero[u] - mid + 1) / 2 * c[u]                      + ans - zero[u];             else                 sum = one * (ans - mid + 1) * b[u]                      + one * (mid + ans) * (ans - mid + 1) / 2 * c[u];         }           if (one * a[u] <= sum)         {             d[u] = mid;             l = mid + 1;         }         else             r = mid - 1;     }     return ; }   int fa[N + 5], in[N + 5]; void dfs(int u, int father) {     fa[u] = father;     in[father]++;     for (int i = head[u]; i; i = nxt[i])     {         int v = to[i];         if (v == father)             continue;         dfs(v, u);     }     return ; }   priority_queue<pir> q; int seq[N + 5]; bool check(long long ans) {     for (int i = 1; i <= n; i++)     {         calc_d(i, ans);         if (d[i] < 0)             return false;     }       dfs(1, 0);     for (int i = 1; i <= n; i++)     {         if (in[i] == 0)             q.emplace(d[i], i);     }       for (int T = n; T > 0; T--)     {         int u = q.top().second;         q.pop();         seq[T] = u;           if (fa[u])         {             in[fa[u]]--;             if (in[fa[u]] == 0)                 q.emplace(d[fa[u]], fa[u]);         }     }       for (int i = 1; i <= n; i++)     {         if (1ll * i > d[seq[i]])             return false;     }     return true; }   int main() {     // freopen("tree.in", "r", stdin);     // freopen("tree.out", "w", stdout);       scanf("%d", &n);     for (int i = 1; i <= n; i++)     {         scanf("%lld%lld%lld", a + i, b + i, c + i);         if (c[i] < 0)             zero[i] = (1ll - b[i]) / c[i];     }     for (int i = 1, u, v; i < n; i++)     {         scanf("%d%d", &u, &v);         add(u, v);     }       long long l = 1, r = Max, ans = 0;     while (l <= r)     {         long long mid = (l + r) >> 1;         if (check(mid))         {             ans = mid;             r = mid - 1;         }         else             l = mid + 1;     }     printf("%lld\n", ans);     return 0; }