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Bzoj3124: [sdoi2013] diameter

2020-12-07 14:26:09 osc_ cl1ufvfd

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solution :
Two silly conclusions :
First of all, it must be on the same diameter . Because every diameter goes through this section .
It must be a continuous segment .
Because if you have two paragraphs . So how did you get there ( There are only two paths on the tree !)



In this way, it's easy to do .
Whatever diameter .
Determine a path on this diameter . Make this path :
It's impossible at any point to extend a diameter from another branch .
I don't want to draw pictures and describe them directly .
Suppose we find a diameter that is 1->2->5->7
So from 2 Start asking ( because 1 There must be no son , Otherwise it won't be the end of the diameter )
So we can record 1 To 2 Distance of , Assuming that t.
And then look at it 2 Can we walk a distance from other branches t. If you can, then 2 None of the previous edges are at the intersection of all diameters







Asked about 5 Just record 1 To 5 Distance of , Assuming that T.
And then look at it 5 Can you come out of another branch T. If you can, then 5 It's not even good on the side before .
And so on .

Ask once, ask backwards .
We've determined the scope .
Then we can solve the problem .

Code implementation :

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
typedef long long ll;
struct node {
  
  int x,y,next;ll d;}a[510000];int len,last[210000];
void ins(int x,int y,ll d) {len++;a[len].x=x;a[len].y=y;a[len].d=d;a[len].next=last[x];last[x]=len;}
int G,fa[210000],bian[210000];ll D;
void dfs(int x,ll d,int f) {
    if(D<d) {D=d;G=x;}
    for(int k=last[x];k;k=a[k].next) {
        int y=a[k].y;
        if(y!=f){fa[y]=x;bian[y]=k;dfs(y,d+a[k].d,x);}
    }
}
int s[210000];
ll T;bool v[210000];
void dfs1(int x,ll d,int f) {
    if(T<d)T=d;
    for(int k=last[x];k;k=a[k].next) {
        int y=a[k].y;
        if(v[y]==true&&y!=f) dfs1(y,d+a[k].d,x);
    }
}
int ans;
void dfs2(int x,int d,int f,int ed) {
    if(x==ed){ans=d;return ;}
    for(int k=last[x];k;k=a[k].next) {
        int y=a[k].y;
        if(y!=f) dfs2(y,d+1,x,ed);
    }
}
int main() {
    int n;scanf("%d",&n);len=0;memset(last,0,sizeof(last));
    for(int i=1;i<n;i++) {
        int x,y;ll d;scanf("%d%d%lld",&x,&y,&d);
        ins(x,y,d);ins(y,x,d);
    }
    int X,Y;
    memset(bian,0,sizeof(bian));dfs(1,0,0);X=G;D=0;dfs(X,0,0);Y=G;
    int x=Y,slen=0;memset(v,true,sizeof(v));
    while(x!=X) {s[++slen]=x;v[x]=false;x=fa[x];}
    s[++slen]=X;v[X]=false;
    int ansx=Y;ll K=a[bian[s[1]]].d;
    for(int i=2;i<=slen;i++) {
        T=0;dfs1(s[i],0,0);
        if(T==K)ansx=s[i];K+=a[bian[s[i]]].d;
    }
    for(int i=slen;i>=2;i--) bian[s[i]]=bian[s[i-1]];bian[s[1]]=0;
    int ansy=X;K=a[bian[s[slen]]].d;
    for(int i=slen-1;i>=1;i--) {
        T=0;dfs1(s[i],0,0);
        if(T==K)ansy=s[i];K+=a[bian[s[i]]].d;
    }dfs2(ansx,0,0,ansy);printf("%lld\n%d\n",D,ans);
    return 0;
}

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