• hdu 2196 Computer 树形DP


    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2196

    A school bought the first computer some time ago(so this computer's id is 1). During the recent years the school bought N-1 new computers. Each new computer was connected to one of settled earlier. Managers of school are anxious about slow functioning of the net and want to know the maximum distance Si for which i-th computer needs to send signal (i.e. length of cable to the most distant computer). You need to provide this information. 

    Hint: the example input is corresponding to this graph. And from the graph, you can see that the computer 4 is farthest one from 1, so S1 = 3. Computer 4 and 5 are the farthest ones from 2, so S2 = 2. Computer 5 is the farthest one from 3, so S3 = 3. we also get S4 = 4, S5 = 4.

    题意描述:给出一颗树,求出每个节点到叶节点的最长距离。

    算法分析:树形DP。每个节点向下dfs一遍,向上dfs一遍即可。

     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cstring>
     4 #include<cstdlib>
     5 #include<cmath>
     6 #include<algorithm>
     7 #include<queue>
     8 #include<vector>
     9 #define inf 0x7fffffff
    10 using namespace std;
    11 const int maxn=10000+10;
    12 
    13 int n;
    14 int fmaxn[maxn],fmaxnid[maxn],smaxn[maxn],smaxnid[maxn];
    15 vector<pair<int,int> >vec[maxn];
    16 
    17 void dfs(int u,int pre)
    18 {
    19     int k=vec[u].size();
    20     for (int i=0 ;i<k ;i++)
    21     {
    22         int v=vec[u][i].first;
    23         int w=vec[u][i].second;
    24         if (v==pre) continue;
    25         dfs(v,u);
    26         if (smaxn[u]<fmaxn[v]+w)
    27         {
    28             smaxn[u]=fmaxn[v]+w;
    29             smaxnid[u]=v;
    30             if (smaxn[u]>fmaxn[u])
    31             {
    32                 swap(fmaxn[u],smaxn[u]);
    33                 swap(fmaxnid[u],smaxnid[u]);
    34             }
    35         }
    36     }
    37     return;
    38 }
    39 
    40 void dfs2(int u,int pre)
    41 {
    42     int k=vec[u].size();
    43     for (int i=0 ;i<k ;i++)
    44     {
    45         int v=vec[u][i].first;
    46         int w=vec[u][i].second;
    47         if (v==pre) continue;
    48         if (v==fmaxnid[u])
    49         {
    50             if (smaxn[u]+w>smaxn[v])
    51             {
    52                 smaxn[v]=smaxn[u]+w;
    53                 smaxnid[v]=u;
    54                 if (smaxn[v]>fmaxn[v])
    55                 {
    56                     swap(smaxn[v],fmaxn[v]);
    57                     swap(smaxnid[v],fmaxnid[v]);
    58                 }
    59             }
    60         }
    61         else if (fmaxn[u]+w>smaxn[v])
    62         {
    63             smaxn[v]=fmaxn[u]+w;
    64             smaxnid[v]=u;
    65             if (smaxn[v]>fmaxn[v])
    66             {
    67                 swap(smaxn[v],fmaxn[v]);
    68                 swap(smaxnid[v],fmaxnid[v]);
    69             }
    70         }
    71         dfs2(v,u);
    72     }
    73     return ;
    74 }
    75 
    76 int main()
    77 {
    78     while (scanf("%d",&n)!=EOF)
    79     {
    80         for (int i=1 ;i<=n ;i++) vec[i].clear();
    81         memset(fmaxn,0,sizeof(fmaxn));
    82         memset(fmaxnid,0,sizeof(fmaxnid));
    83         memset(smaxn,0,sizeof(smaxn));
    84         memset(smaxnid,0,sizeof(smaxnid));
    85         int a,b;
    86         for (int i=2 ;i<=n ;i++)
    87         {
    88             scanf("%d%d",&a,&b);
    89             vec[i].push_back(make_pair(a,b));
    90             vec[a].push_back(make_pair(i,b));
    91         }
    92         dfs(1,1);
    93         dfs2(1,1);
    94         for (int i=1 ;i<=n ;i++) printf("%d
    ",fmaxn[i]);
    95     }
    96     return 0;
    97 }
  • 相关阅读:
    java 内存命令
    sts4 集成 springboot 和 activiti5
    sts4 安装spring xml 的提示。
    去除office 2019 由于您的登录凭据已经过期,的提示。
    「なきゃ」和「なくちゃ」分别是什么的原型?
    能力考必看|推荐三本适合N1和N2能力考练听力的书|帝京日语
    影子练习
    早大前辈分享日本留学生考试(EJU)日语高分经验
    星野源 新垣结衣 结婚
    使Nginx支持更多并发 请求的 Linux内核参数的优化
  • 原文地址:https://www.cnblogs.com/huangxf/p/4370159.html
Copyright © 2020-2023  润新知