zoukankan      html  css  js  c++  java
  • (树形DP) hdu 2196

    Computer

    Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 3982    Accepted Submission(s): 1996


    Problem Description
    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.
     
    Input
    Input file contains multiple test cases.In each case there is natural number N (N<=10000) in the first line, followed by (N-1) lines with descriptions of computers. i-th line contains two natural numbers - number of computer, to which i-th computer is connected and length of cable used for connection. Total length of cable does not exceed 10^9. Numbers in lines of input are separated by a space.
     
    Output
    For each case output N lines. i-th line must contain number Si for i-th computer (1<=i<=N).
     
    Sample Input
    5 1 1 2 1 3 1 1 1
     
    Sample Output
    3 2 3 4 4
     

    三次dfs ORZ

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<string>
    #include<cmath>
    #include<cstdlib>
    #include<algorithm>
    #include<vector>
    #include<set>
    #include<stack>
    #include<stack>
    using namespace std;
    int n,ed,maxx;
    struct node
    {
        int v,w;
        node(int _v,int _w) : v(_v),w(_w) {}
    };
    vector<node> e[10005];
    int dist[10005];
    void dfs(int u,int father,int len)
    {
        if(len>maxx) maxx=len,ed=u;
        for(int i=0;i<e[u].size();i++)
        {
            int v=e[u][i].v;
            int w=e[u][i].w;
            if(v==father)
                continue;
            dfs(v,u,len+w);
            dist[v]=max(dist[v],len+w);
        }
    }
    int main()
    {
        while(scanf("%d",&n)!=EOF)
        {
            maxx=0;
            memset(dist,0,sizeof(dist));
            for(int i=1;i<=n;i++)
                e[i].clear();
            for(int i=2;i<=n;i++)
            {
                int x,y;
                scanf("%d%d",&x,&y);
                e[i].push_back(node(x,y));
                e[x].push_back(node(i,y));
            }
            dfs(1,-1,0);
            dfs(ed,-1,0);
            dfs(ed,-1,0);
            for(int i=1;i<=n;i++)
                printf("%d
    ",dist[i]);
        }
        return 0;
    }
    

      

  • 相关阅读:
    app移动测试 (自动化遍历方法和技巧)(转载大佬)
    使用Jmeter录制脚本并调试
    整理token,session ,cookies 和正则表达式整理
    ·接口测试核心:URL&HTTP协议详解
    性能测试基本概念 聚合报告指标分析
    数据分析技能点梳理
    数据分析整体知识点架构(转载)
    数据分析职业前景规划
    数据分析三年建议指导(书籍等)
    系统吞吐量(TPS)、用户并发量、性能测试概念和公式 (转载)
  • 原文地址:https://www.cnblogs.com/water-full/p/4493844.html
Copyright © 2011-2022 走看看