Original link: https://www.shuizilong.com/house/archives/saco-2018-february-contest-gold-problem-2-directory-traversal/
Basically Tree Distances II,
The transfer function is adjusted slightly.
#include <lastweapon/io> using namespace lastweapon; const int N = int(2e5) + 9; int n, m, h[N], c[N], l[N]; LL f[N], g[N]; VVI v; void dfs1(int r, int p = -1) { for(int& a : v[r]) if (a != p) { dfs1(a, r); h[r] += h[a]; f[r] += f[a] + h[a]*l[a]; } if(!c[r]) h[r]++; } void dfs2(int r, int p = -1) { for(int& a : v[r]) if (a != p) { g[a] = (f[r] - (f[a] + h[a]*l[a])) + g[r] + (m - h[a])*3; dfs2(a, r); } } int main() { freopen("dirtraverse.in", "r", stdin); freopen("dirtraverse.out", "w", stdout); RD(n); v. resize(n); int a; string s; REP(i, n) { cin >> s; l[i] = s. size(); RD(c[i]); if(c[i]) l[i]++; else m++; REP(j, c[i]) { RD(a), a--; v[i].PB(a); v[a].PB(i); } } dfs1(0); dfs2(0); LL ans = INFF; REP(i, n) ans = min(ans, f[i]+g[i]); cout << ans << endl; }
This article is reproduced from: https://www.shuizilong.com/house/archives/saco-2018-february-contest-gold-problem-2-directory-traversal/
This site is only for collection, and the copyright belongs to the original author.