xtu 1267 Highway 湘潭邀请赛H

最新推荐文章于 2020-11-07 16:53:26 发布

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最新推荐文章于 2020-11-07 16:53:26 发布

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分类专栏： c/c++

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Highway

In ICPCCamp there were n towns conveniently numbered with 1,2,…,n connected with (n−1) roads. The i -th road connecting towns ai and bi has length ci . It is guaranteed that any two cities reach each other using only roads.

Bobo would like to build (n−1) highways so that any two towns reach each using only highways. Building a highway between towns x and y costs him δ(x,y) cents, where δ(x,y) is the length of the shortest path between towns x and y using roads.

As Bobo is rich, he would like to find the most expensive way to build the (n−1) highways.

Input

The input contains zero or more test cases and is terminated by end-of-file. For each test case:

The first line contains an integer n . The i -th of the following (n−1) lines contains three integers ai , bi and ci .

1≤n≤105
1≤ai,bi≤n
1≤ci≤108
The number of test cases does not exceed 10 .

Output

For each test case, output an integer which denotes the result.

Sample Input

Sample Output

19
15

http://www.dengwenhuo.cn/?id=453

#include <stdio.h>
#include <algorithm>
#include <vector>
#include <string.h>
#include <queue>
using namespace std;
 
#define ll __int64
const int N=100005;
 
struct p{
    ll to,val;
};
 
struct pp{
    ll u,dis;
    bool operator < (const pp&r) const{
        return dis>r.dis;
    }
};
vector<p>G[N];
int n,vis[N];
ll dis[N];
 
int dijkstra(int s)
{
    memset(vis,0,sizeof(vis));
    memset(dis,0x3f,sizeof(dis));
    priority_queue<pp>q;
    dis[s]=0;
    q.push(pp{s,0});
    int k=1;
    while(!q.empty())
    {
        pp now=q.top();
        q.pop();
        if(vis[now.u]) continue;
        vis[now.u]=1;
        for(int i=0;i<G[now.u].size();i++)
        {
            p t=G[now.u][i];
            if(dis[t.to]>dis[now.u]+t.val)
            {
                dis[t.to]=dis[now.u]+t.val;
                if(dis[k]<dis[t.to])
                    k=t.to;
                q.push(pp{t.to,dis[t.to]});
            }
        }
    }
    return k;
}
void get(int s)
{
    memset(vis,0,sizeof(vis));
    queue<pp>q;
    vis[s]=1;
    q.push(pp{s,0});
    while(!q.empty())
    {
        pp now=q.front();
        q.pop();
        dis[now.u]=max(now.dis,dis[now.u]);
        for(int i=0;i<G[now.u].size();i++)
        {
            p t=G[now.u][i];
            if(!vis[t.to])
            {
                vis[t.to]=1;
                q.push(pp{t.to,now.dis+t.val});
            }
        }
    }
}
int main()
{
    ll u,v,c;
    int k1,k2;
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++) G[i].clear();
        for(int i=1;i<n;i++)
        {
            scanf("%I64d %I64d %I64d",&u,&v,&c);
            G[u].push_back(p{v,c});
            G[v].push_back(p{u,c});
        }
        k1=dijkstra(1);
        ll t=dis[k1];
        k2=dijkstra(k1);
        t=max(t,dis[k2]);
        memset(dis,0,sizeof(dis));
        get(k1);get(k2);
        ll ans=0;
        for(int i=1;i<=n;i++)
            ans+=dis[i];
        printf("%I64d\n",ans-t);
    }
    return 0;
}