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  • 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 (n1) 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 (n1) 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 (n1) 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 (n1) lines contains three integers ai , bi and ci .

    • 1n105
    • 1ai,bin
    • 1ci108
    • 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
    ",sum-Max);
    80     }
    81     return 0;
    82 }
     
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  • 原文地址:https://www.cnblogs.com/yinghualuowu/p/6875605.html
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