P3304 [SDOI2013] 直径

合集 - 模拟(54)

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54.P3304 [SDOI2013] 直径2024-07-15

收起

原题链接

题解

先随便找一条直径，然后标记这些边，然后看看直径上的点有没有不需要经过标记边的路径，使得其长度等于该点到直径端点的路径长度

code

#include<bits/stdc++.h>
#define ll long long
using namespace std;

struct edge
{
    ll to,val;
};
vector<edge> G[200005];

ll maxh=0;
ll node=1;

void dfs(ll now,ll fa,ll height)
{
    if(height>maxh)
    {
        maxh=height;
        node=now;
    }

    for(auto next:G[now])
    {
        if(next.to==fa) continue;

        dfs(next.to,now,height+next.val);
    }
}

vector<ll> chain;
ll pre[200005]={0};
bool vis[200005]={0};

bool query(ll now,ll fa,ll des)
{
    if(now==des)
    {
        chain.push_back(now);
        vis[now]=1;
        return 1;
    }

    for(auto next:G[now])
    {
        ll to=next.to;
        if(to==fa) continue;

        if(query(to,now,des))
        {
            chain.push_back(now);
            ll tem=chain.size()-1;
            pre[tem]=pre[tem-1]+next.val;
            vis[now]=1;
            return 1;
        }
    }
    return 0;
}

ll check(ll now)
{
    ll res=0;
    for(auto next:G[now])
    {
        ll to=next.to,val=next.val;
        if(vis[to]) continue;
        vis[to]=1;
        res=max(res,check(to)+val);
        vis[to]=0;
    }
    return res;
}

void solve()
{
    ll n;
    cin>>n;

    for(ll i=1;i<n;i++)
    {
        ll x,y,w;
        cin>>x>>y>>w;
        G[x].push_back({y,w});
        G[y].push_back({x,w});
    }

    dfs(1,1,0);
    ll node1=node;

    maxh=0;
    node=node1;
    dfs(node1,node1,0);
    ll node2=node;

    query(node1,node1,node2);

    ll len=chain.size();
    ll l=0,r=len-1;

    for(ll i=1;i<len-1;i++)
    {
        ll res=check(chain[i]);
        ll sum=pre[i];
        if(res==sum) l=i;
        if(res==pre[len-1]-sum)
        {
            r=i;
            break;
        }
    }

    cout<<pre[len-1]<<'\n';
    cout<<r-l<<'\n';
}
int main()
{
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    ll t=1;
    while(t--) solve();
    return 0;
}