hdu 2196 Computer（樹形圖+bfs）
來源：程序員人生 發布時間：2015-07-24 09:07:03 閱讀次數：3919次
Computer
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3956 Accepted Submission(s): 1983
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 N⑴ 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 (N⑴) 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
Sample Output
求樹上1點到其它點的最遠距離，先以任1點為根節點，進行1次bfs搜到距離當前點最遠的點root1，從root1開始再搜，得到root2，從root2再搜1次得到終究答案，每次搜的進程中更新每一個點的最遠距離。
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
#define ll long long
#define N 11000
#define mem(a,t) memset(a,t,sizeof(a))
struct node
{
int v,w,next;
}e[N*2];
int head[N];
int dis[N];
int cnt;
int len_max,root;
void add(int u,int v,int w)
{
e[cnt].v=v;
e[cnt].w=w;
e[cnt].next=head[u];
head[u]=cnt++;
}
void bfs(int u,int len,int fa)
{
int i,v;
if(len>len_max)
{
len_max=len;
root=u;
}
for(i=head[u];i!=⑴;i=e[i].next)
{
v=e[i].v;
if(v!=fa)
{
if(dis[v]<len+e[i].w)
dis[v]=len+e[i].w;
bfs(v,len+e[i].w,u);
}
}
return ;
}
int main()
{
int i,n,a,b;
while(~scanf("%d",&n))
{
mem(head,⑴);
cnt=0;
for(i=2;i<=n;i++)
{
scanf("%d%d",&a,&b);
add(i,a,b);
add(a,i,b);
}
mem(dis,0);
len_max=0;
bfs(1,0,⑴);
len_max=0;
bfs(root,0,⑴);
bfs(root,0,⑴);
for(i=1;i<=n;i++)
printf("%d
",dis[i]);
}
return 0;
}
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