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

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

题目大意：给你一颗已知的树，求出新建的一颗权值最大的树的权值

解题思路：树形dp，求出每个点到树其他点的最长距离再减去树中的最长路径。

将每个点的最长路径添加到新建的树中，除了树中最长的那条路径会重复一次（构成一个环），不会形成其他环，而所有的点都添加到了新建的树中，于是构成了一颗权值最大的树。

dp[u][0]:u到子树节点的最长距离，dp[u][1]:u到子树节点外其他节点的最长距离

dp[u][0]很容易在dfs中计算出来。

dp[v][1]:若v不是u子树最长路径经过的点，dp[v][1] = max( dp[u][1]+Luv , dp[u][0]+Luv )

若v是u子树最长路径所经过的点，dp[v][1] = max( dp[u][1]+Luv , dp[u][0]+Se[u])

Se[u]带表u的子树中第二长的路径长度。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <queue>
#include <set>
#include <string>
#include <stack>
#include <algorithm>
#include <map>
using namespace std;
typedef  __int64 ll;
const int N = 100100;
const int M = 20;
const int INF = 0x3fffffff;struct  Edge
{int node,len;Edge*next;
}m_edge[N*2];
int Ecnt;
Edge*head[N];
int vis[N],flag[N];
ll dp[N][2],second[N];
//dp[u][0]:u到子树节点的最长距离,dp[u][1]:u到子树节点外其他节点的最长距离
//second[u]:子树节点到u的第二长路径，flag[v]:节点v的上层节点的最长路径是否经过vvoid init()
{Ecnt = 0;fill( head , head+N , (Edge*)0 );fill( flag , flag+N , 0 );fill( second , second+N , 0 );
}void mkEdge( int a , int b , int c )
{m_edge[Ecnt].node = a;m_edge[Ecnt].len = c;m_edge[Ecnt].next = head[b];head[b] = m_edge+Ecnt++;m_edge[Ecnt].node = b;m_edge[Ecnt].len = c;m_edge[Ecnt].next = head[a];head[a] = m_edge+Ecnt++;
}ll dfs1( int u )
{vis[u] = true; dp[u][0] = 0;for( Edge*p = head[u] ; p ; p = p->next ){int v = p->node;if( !vis[v] ){ll w = dfs1( v );dp[u][0] = max( dp[u][0] , w+p->len );}}return dp[u][0];
}void dfs2( int u  )
{vis[u] = true;int fg = 0;for( Edge*p = head[u] ; p ; p = p ->next ){int v = p->node;if( !vis[v] ){if( !fg && dp[u][0] == dp[v][0]+p->len ) { flag[v] = 1;fg = 1; }else second[u] = max( second[u] , dp[v][0]+p->len );dfs2( v );}}
}void dfs3( int u )
{vis[u] = true;for( Edge*p = head[u] ; p ; p = p->next ){int v = p->node;if( !vis[v] ){if( !flag[v] ) dp[v][1] = max( dp[u][1]+p->len , dp[u][0]+p->len );else dp[v][1] = max( dp[u][1]+p->len , second[u]+p->len );dfs3( v );}}
}int main()
{int n,a,b,c;while( ~scanf("%d",&n) ){init();for( int i = 0 ; i < n-1 ; ++i ){scanf("%d%d%d",&a,&b,&c);mkEdge( a , b , c );}memset( vis , 0 , sizeof(vis) );dfs1( 1 );memset( vis , 0 , sizeof(vis) );dfs2( 1 );memset( vis , 0 , sizeof(vis) );dfs3( 1 );ll rec = 0,ans = 0;for( int i = 1 ; i <= n ; ++i ){ll t = max( dp[i][0] , dp[i][1] );ans += t;rec = max( rec , t );}ans -= rec;printf("%I64d\n",ans);}return 0;
}

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