2017年“嘉杰信息杯” 中国大学生程序设计竞赛全国邀请赛 Highway

Highway
Accepted : 122   Submit : 393
Time 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

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

Source

XTU OnlineJudge 
 题意：要把所有的边都联通，并要求权值之和最大
 解法：因为是颗树，那就没必要讨论两点的最短路了（毕竟树是没有回路的），那么重点是落在了如何找到权值之和最大的方法
 我们找到这颗树相距最远的两个点（树的直径）A，B 剩余的点取距离A或者B最远的那条，需要进行两次dfs，距离保存最大的
 然后加起来就行，因为A到B我们多加了一次，再减去就是结果
 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 const int inf=(1<<30);
 5 const int maxn=100005;
 6 ll pos;
 7 ll n,ans,vis[maxn],in[maxn];
 8 vector<pair<int,int>>e[maxn];
 9 ll sum;
10 void dfs(int v,ll cnt)
11 {
12     if(ans<cnt)
13     {
14         ans=cnt;
15         pos=v;
16     }
17     if(vis[v])return;
18     vis[v]=1;
19     for(int i=0; i<e[v].size(); i++)
20         //    cout<<e[v][i].first;
21         if(!vis[e[v][i].first])
22             dfs(e[v][i].first,cnt+e[v][i].second);
23 }
24 ll dis1[123456],dis2[123456];
25 void DFS(int v,ll cnt,ll dis[])
26 {
27     if(vis[v]) return;
28     vis[v]=1;
29     dis[v]=cnt;
30     for(int i=0; i<e[v].size(); i++)
31         //    cout<<e[v][i].first;
32         if(!vis[e[v][i].first])
33             DFS(e[v][i].first,cnt+e[v][i].second,dis);
34 }
35 int main()
36 {
37     int n,m;
38     ans=0;
39     while(~scanf("%d",&n))
40     {
41         ans=0;
42         memset(dis1,0,sizeof(dis1));
43         memset(dis2,0,sizeof(dis2));
44         memset(in,0,sizeof(in));
45         memset(vis,0,sizeof(vis));
46         for(int i=0;i<=n;i++)
47         {
48             e[i].clear();
49         }
50         for(int i=1; i<n; i++)
51         {
52             ll u,v,w;
53             scanf("%d%d%d",&u,&v,&w);
54             e[u].push_back({v,w});
55             e[v].push_back({u,w});
56         }
57         dfs(1,0);
58         ll cnt=ans;
59         ans=0;
60         memset(vis,0,sizeof(vis));
61         ans=0;
62         DFS(pos,0,dis1);
63         memset(vis,0,sizeof(vis));
64         ans=0;
65         dfs(pos,0);
66 
67         memset(vis,0,sizeof(vis));
68         DFS(pos,0,dis2);
69         memset(vis,0,sizeof(vis));
70         ll cot=ans;
71         //cout<<cot<<" "<<cnt<<endl;
72         ll Max=max(cnt,cot);
73         //cout<<Max<<endl;
74         sum=0;
75         for(int i=1;i<=n;i++)
76         {
77             sum+=max((ll)dis1[i],(ll)dis2[i]);
78         }
79         printf("%lld\n",sum-Max);
80     }
81     return 0;
82 }

 

 
posted @ 2017-05-18 21:07  樱花落舞  阅读(320)  评论(0)  编辑  收藏  举报