今日SGU 5.12

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;
}

原文地址：https://www.cnblogs.com/chinacwj/p/9030061.html