江苏省赛--湘潭杯-highway 题解

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

5
1 2 2
1 3 1
2 4 2
3 5 1
5
1 2 2
1 4 1
3 4 1
4 5 2
Sample Output

19
15
题目链接： http://202.197.224.59/OnlineJudge2/index.php/Problem/index/p/13/
或 https://www.icpc-camp.org/contests/4mYguiUR8k0GKE
题目大意：
给出一棵无根树，给出n-1条边和权值，然后让你求重新选择n-1条边使其联通并且权和最大
官方题解：

首先有一点得知道，关于树的直径
树的直径是指树上权值和最大的路径（最简单路径，即每一个点只经过一次）
存在结论：对于树上的任意一个节点，距离这个节点最远的距离一定是到直径的端点的距离

就是先求出树的最远点对(树的直径的端点)d1,d2，再求出以直径的两个端点为起点的disi，首先将直径(d1,d2的距离)加入集合，对于其他点i，加入max(d1到i的距离,d2到i的距离)到集合，集合所构成的树就是题目的答案

#include <iostream>#include<algorithm>#include<string>#include<cstring>#include<vector>#include<cstdlib>#include<queue>#include<cstdio>using namespace std;const int maxn = 100100;struct node{    int u,v,w;    int next;} e[maxn<<1];long long dis[maxn],dd[maxn];int head[maxn];int top;bool vis[maxn];void init(){    memset(dis,0,sizeof(dis));    memset(head,-1,sizeof(head));    memset(vis,false,sizeof(vis));    top = 0;}void add(int u,int v,int w){    e[top].u = u;    e[top].v = v;    e[top].w = w;    e[top].next = head[u];    head[u] = top++;}int bfs(int s)//求树的直径{    queue<int>q;    vis[s] = true;    q.push(s);    int key = - 1;     long long mmax = 0;    while(!q.empty())    {        int t = q.front();        q.pop();        for(int i=head[t]; ~i; i=e[i].next)        {            int v = e[i].v;//当前思想，把这个点看成结果理解下面的if            if(!vis[v]&&dis[v]<dis[t]+e[i].w)//最大路判断            {                vis[v] = true;                dis[v] = dis[t]+e[i].w;                q.push(v);                if(dis[v]>mmax)                {                    mmax = dis[v];                    key = v;                }            }        }    }    return key;}void bfs1(int s){    queue<int>q;    vis[s] = true;    q.push(s);    while(!q.empty())    {        int t = q.front();        q.pop();        for(int i=head[t]; ~i; i =e[i].next)        {            int v = e[i].v;            if(!vis[v]&&dis[v]<dis[t]+e[i].w)            {                vis[v] =true;                dis[v] = dis[t]+e[i].w;                dd[v] = max(dis[v],dd[v]);                q.push(v);            }        }    }}int main(){    int n,u,v,w;    while(cin>>n)    {        init();        memset(dd,0,sizeof(dd));        for(int i=0; i<n-1; i++)        {            scanf("%d%d%d",&u,&v,&w);            add(u,v,w);            add(v,u,w);        }        int a = bfs(1);        memset(vis,false,sizeof(vis));        memset(dis,0,sizeof(dis));        int b = bfs(a);        long long sum = dis[b];        memset(vis,false,sizeof(vis));        memset(dis,0,sizeof(dis));        bfs1(a);        memset(vis,false,sizeof(vis));        memset(dis,0,sizeof(dis));        bfs1(b);        for(int i=1; i<=n; i++)        {            if(i==a||i==b)                continue;            sum+=dd[i];        }        cout<<sum<<endl;    }    return 0;}

参考：
http://blog.csdn.net/bless924295/article/details/72331272
http://blog.csdn.net/liangzhaoyang1/article/details/72796738