HDU-2196 Computer

D - Computer

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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Status
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

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1. 题意：给你多个数据，比如上例：5代表有5个节点；第二行代表第二个节点和第一个节点相连，且边权是1；第三行代表第三个节点何地二个节点相连接，边权为1.
2. 思路：根据题所给的信息每个新电脑和原来的电脑连接可以判断所给图形为一棵树，题意就是要求树的每一点与其他点的最大距离，可以借助树的直径进行转化（到树上任一点的最大距离的点一定树的直径上而且是两端）：任选一点跑一遍bfs找到直径的的起点S，再用S点为起点跑一遍bfs找到直径的终点E,并保存此时每个点和S的距离；再用E为起点跑一遍bfs，记录每个点与其的距离。再比较两次结果，取最大者。
3. 失误：刚开始没用树的直径，循环一遍bfs超时，感觉应该用树的直径做，但写出来测试数据对，提交错误。看了学长的思路才明白，我把任意的点都默认成直径上的点了，就是按一条线段上的点一样考虑了；改正之后还是有些小错误，在代码中指出。
4. 代码如下：

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#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;

const int MAXN=1e6+10;
int head[MAXN],dis[MAXN],d1[MAXN],d2[MAXN];
bool vis[MAXN];
int edgenum,ans,Tnode,n;

struct Edge{
    int from;
    int to; 
    int val;
    int next;

}edge[MAXN];

void init()
{
    memset(head,-1,sizeof(head)); edgenum=0;  
}

void addedge(int u,int v,int w)
{
    Edge E={u,v,w,head[u]};//用这种方式初始化结构体中变量顺序不能颠倒  
    edge[edgenum]=E;   //错误现象：死机状态 
    head[u]=edgenum++;
}

void bfs(int s)
{
    memset(dis,0,sizeof(dis));  memset(vis,false,sizeof(vis));
    queue<int> que;  que.push(s); dis[s]=0; vis[s]=true; ans=0;
    while(!que.empty()) 
    {
        int now=que.front(); que.pop();
        for(int i=head[now];~i;i=edge[i].next)
        {
              int go=edge[i].to;
              if(!vis[go])
              {
                  if(dis[go]<dis[now]+edge[i].val)  dis[go]=dis[now]+edge[i].val;
//                if(ans<dis[go])//刚学邻接表，不怎麽理解，先记着用的:dis保存的是数组下标所带表的节点到起始节点的距离
//                {
//                     ans=dis[go];
//                     Tnode=go;
//                  }
                    vis[go]=true;
                    que.push(go);
              }
         } 
    }
    for(int i=1;i<=n;++i)//跑的和在里面差不多 
    {
        if(ans<dis[i])
        {
            ans=dis[i];
            Tnode=i;
        }
    }
}

int main()
{
    int i,v,w,S,end;
    while(~scanf("%d",&n))
    {
        init();
        for(i=2;i<=n;++i)//没想到i从2开始 
        {
            scanf("%d%d",&v,&w);
            addedge(i,v,w);
            addedge(v,i,w);
        }
        bfs(1);  S=Tnode;
        bfs(S);  end=Tnode;
        for(i=1;i<=n;++i)  d1[i]=dis[i] ;
        bfs(end); 
    //  for(i=1;i<=n;++i)  d2[i]=dis[i];
        for(i=1;i<=n;++i)//跑的快了三分之一 
        {
            ans=d1[i]>dis[i]?d1[i]:dis[i];
            printf("%d\n",ans); 
         } 
    }
    return 0;
 }