2017湘潭赛XTU1267Highway

Highway

Accepted : 169 Submit : 577Time Limit : 4000 MS Memory Limit : 65536 KB

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

51 2 21 3 12 4 23 5 151 2 21 4 13 4 14 5 2

Sample Output

1915

Source

XTU OnlineJudge

http://202.197.224.59/OnlineJudge2/index.php/Problem/read/id/1267

本题的题意是真的坑爹   告诉你一颗树 你就可以求出两两的距离  再用这些距离去构造最大生成树  4到5的距离为6 

此题需要知道一个东西 就是树的直径   http://www.cnblogs.com/xubenben/archive/2012/12/28/2837971.html

#include<stdio.h>#include<vector>#include<queue>#include<string.h>using namespace std;#define ll long longbool vis[100010];ll gg[100010];ll gg1[2][100010];int pp;ll total;int qd;struct node{    int yd,val,et;};vector<node>ap[100010];vector<node>ap1[100010];int bfs(int ss){    int zd;    queue<int>qe;    qe.push(ss);    vis[ss]=1;    ll  maxx=0;    while(!qe.empty())    {        int dt=qe.front();//原点        qe.pop();        for(int i=0; i<ap[dt].size(); i++)        {            node  ft=ap[dt][i];            if(vis[ft.et]) continue;            vis[ft.et]=1;            gg[ft.et]=gg[dt]+ap[dt][i].val;            qe.push(ft.et);            if(maxx<gg[ft.et])            {                maxx=gg[ft.et];                zd=ft.et;            }        }    }    return zd;}void dfs(int k,ll sum){    for(int i=0; i<ap1[k].size(); i++)    {        int vv=ap1[k][i].et;        if(vis[vv]) continue;        vis[vv]=1;        dfs(vv,sum+ap1[k][i].val);    }    if(total<sum)    {        total=sum;        qd=k;    }}void dfs1(int k,ll sum){    for(int i=0; i<ap1[k].size(); i++)    {        int vv=ap1[k][i].et;        if(vis[vv]) continue;        vis[vv]=1;        gg1[pp][vv]=sum+ap1[k][i].val;        dfs1(vv,sum+ap1[k][i].val);    }}int main(){    int n,i;    while(scanf("%d",&n)!=EOF)    {        for(i=0; i<=n; i++)        {            ap[i].clear();            ap1[i].clear();        }        node xf;        int u,v,w;        for(i=1; i<n; i++)        {            scanf("%d %d %d",&u,&v,&w);            xf.yd=u;            xf.et=v;            xf.val=w;            ap[u].push_back(xf);            ap1[u].push_back(xf);            xf.yd=v;            xf.et=u;            ap[v].push_back(xf);            ap1[v].push_back(xf);        }        memset(vis,0,sizeof(vis));        int zd=bfs(1); //求直径一端点        memset(vis,0,sizeof(vis));        total=0;        vis[zd]=1;        dfs(zd,0);   //求直径长度        // printf("%d %d %I64d\n",qd,zd,total);                //求到两个端点的距离        memset(vis,0,sizeof(vis));        vis[zd]=1;        pp=0;        dfs1(zd,0);                memset(vis,0,sizeof(vis));        vis[qd]=1;        pp=1;        dfs1(qd,0);        ll sum=0;        gg1[0][zd]=0;        gg1[1][qd]=0;        for(i=1; i<=n; i++)        {            //printf("%I64d %I64d\n",gg1[0][i],gg1[1][i]);            sum+=max(gg1[0][i],gg1[1][i]);        }        sum-=gg1[0][qd];        printf("%I64d\n",sum);    }    return 0;}