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

In ICPCCamp there were . It is guaranteed that any two cities reach each other using only roads.

Bobo would like to build using roads.

As Bobo is rich, he would like to find the most expensive way to build the 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 .

The number of test cases does not exceed .

Output

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

Sample Input

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 }