luoguP2052 [NOI2011]道路修建

Date: Wed 31 October 2018
Updated: Wed 31 October 2018

In 5. OI.

Tags: OI

题目


O(n)用树上DP或者拓扑跑一遍

拓扑:当度为1时,加入队列,每次f[i]被加时,度--

代码是树上DP的

#include<bits/stdc++.h>
#define N 1000010
#define abs(a) ((a)>0?(a):(-(a)))
#define mem(a,b) memset(a,b,sizeof(a))
#define ll long long
#define fsb(a,b,c) for(ll a=b;a<=c;a++)
#define fbs(a,b,c) for(ll a=b;a>=c;a--)
using namespace std;
struct edge{
 ll p,nt,mark;
}a[N*2];
struct node{
 ll x,y,z;
}b[N];
ll n,cnt=0,head[N],x,y,z,ans=0,f[N];
template<typename qw>inline void rll(qw &x){
 x=0;qw f=1;char c=getchar();
 while(!isdigit(c))f=c=='-'?-1:f,c=getchar();
 while(isdigit(c))x=x*10+c-'0',c=getchar();
 x*=f;
}
inline void add(ll x,ll y,ll z){
 a[++cnt].p=y;a[cnt].nt=head[x];head[x]=cnt;a[cnt].mark=z;
}
inline void dfs(ll u,ll fa){
 f[u]=1;
 for(ll t=head[u];t!=-1;t=a[t].nt){
  ll v=a[t].p;if(v==fa)continue;
  dfs(v,u);f[u]+=f[v];ans+=abs(n-2*f[v])*b[a[t].mark].z;
 }
}
int main(){
 rll(n);mem(head,255);
 fsb(i,1,n-1){
  rll(x);rll(y);rll(z);add(x,y,i);add(y,x,i);
  b[i]=(node){x,y,z};
 }
 mem(f,0);dfs(1,0);
 printf("%lld\n",ans);
 return 0;
}

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