SGU 149
题意：求每一个点的距离最远距离的点的长度
收获：次大值和最大值，dfs
#include<bits/stdc++.h> #define de(x) cout<<#x<<"="<<x<<endl; #define dd(x) cout<<#x<<"="<<x<<" "; #define rep(i,a,b) for(int i=a;i<(b);++i) #define repd(i,a,b) for(int i=a;i>=(b);--i) #define repp(i,a,b,t) for(int i=a;i<(b);i+=t) #define ll long long #define mt(a,b) memset(a,b,sizeof(a)) #define fi first #define se second #define inf 0x3f3f3f3f #define INF 0x3f3f3f3f3f3f3f3f #define pii pair<int,int> #define pdd pair<double,double> #define pdi pair<double,int> #define mp(u,v) make_pair(u,v) #define sz(a) (int)a.size() #define ull unsigned long long #define ll long long #define pb push_back #define PI acos(-1.0) #define qc std::ios::sync_with_stdio(false) #define db double #define all(a) a.begin(),a.end() const int mod = 1e9+7; const int maxn = 1e4+6; const double eps = 1e-6; using namespace std; bool eq(const db &a, const db &b) { return fabs(a - b) < eps; } bool ls(const db &a, const db &b) { return a + eps < b; } bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); } ll gcd(ll a,ll b) { return a==0?b:gcd(b%a,a); }; ll lcm(ll a,ll b) { return a/gcd(a,b)*b; } ll kpow(ll a,ll b) {ll res=1;a%=mod; if(b<0) return 1; for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;} ll read(){ ll x=0,f=1;char ch=getchar(); while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } //inv[1]=1; //for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod; int mx[maxn][2] = {0}; vector<pii> G[maxn]; int dfs(int u,int p){ rep(i,0,sz(G[u])){ int v=G[u][i].se,w=G[u][i].fi; if(v==p) continue; int tmp = dfs(v,u) + w; rep(i,0,2) if(mx[u][i]<tmp) swap(mx[u][i],tmp); } return mx[u][0]; } void dp(int u,int p){ rep(i,0,sz(G[u])){ int v=G[u][i].se,w=G[u][i].fi; if(v==p) continue; int tmp; if(mx[v][0] + w == mx[u][0]) tmp = mx[u][1] + w; else tmp = mx[u][0] + w; rep(i,0,2) if(mx[v][i]<tmp) swap(mx[v][i],tmp); dp(v,u); } } int main(){ int n; scanf("%d",&n); rep(i,2,n+1){ int u,c; scanf("%d%d",&u,&c); G[u].pb(mp(c,i)); G[i].pb(mp(c,u)); } dfs(1,-1);dp(1,-1); rep(i,1,n+1) printf("%d\n",mx[i][0]); return 0; }