Computer
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 35047 Accepted Submission(s):
5633
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.
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 exampleinput 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).
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
题意：给你一棵树，求每个节点的最远对点的距离。
题目看似有点难度，但我们想一想两遍dfs求树的直径的过程，第一遍dfs找的是随意一个点的最远对点，它一定是树的直径的端点，那么我们就可以猜想一下是不是每个点的最远对点都是树的直径的端点呢？答案是肯定的。那么我们就可以先把树的直径找出来然后分别求两个端点到所有的点的距离，取最大值即可。
代码如下：
#include<iostream>
#include<cstdio>
#include<string>
#include<cmath>
#include<cstring>
#include<queue>
#include<stack>
#include<algorithm>
#define maxn 10005
using namespace std; struct edge
{
int next;
int to;
int dis;
}g[maxn<<]; inline int read()
{
char c=getchar();
int res=,x=;
while(c<''||c>'')
{
if(c=='-')
x=-;
c=getchar();
}
while(c>=''&&c<='')
{
res=res*+(c-'');
c=getchar();
}
return x*res;
} int n,aa,bb,num,root,ans;
int last[maxn],d[maxn],dp[maxn],dp1[maxn]; inline void add(int from,int to,int dis)
{
g[++num].next=last[from];
g[num].to=to;
g[num].dis=dis;
last[from]=num;
} void dfs(int x)
{
d[x]=;
for(int i=last[x];i;i=g[i].next)
{
int v=g[i].to;
if(!d[v])
{
dp[v]=dp[x]+g[i].dis;
dfs(v);
}
}
} void dfs1(int x)
{
d[x]=;
for(int i=last[x];i;i=g[i].next)
{
int v=g[i].to;
if(!d[v])
{
dp[v]=dp[x]+g[i].dis;
dfs1(v);
}
}
} void dfs2(int x)
{
d[x]=;
for(int i=last[x];i;i=g[i].next)
{
int v=g[i].to;
if(!d[v])
{
dp1[v]=dp1[x]+g[i].dis;
dfs2(v);
}
}
} int main()
{
while(scanf("%d",&n)!=EOF)
{
ans=;num=;
memset(last,,sizeof(last));
memset(dp,,sizeof(dp));
memset(d,,sizeof(d));
memset(dp1,,sizeof(dp1));
for(int i=;i<=n;i++)
{
aa=read();bb=read();
add(i,aa,bb);
add(aa,i,bb);
}
dfs();
for(int i=;i<=n;i++)
{
if(dp[i]>ans)
{
ans=dp[i];
root=i;
}
}
memset(dp,,sizeof(dp));
memset(d,,sizeof(d));
dfs1(root);
ans=;
memset(d,,sizeof(d));
for(int i=;i<=n;i++)
{
if(dp[i]>ans)
{
ans=dp[i];
root=i;
}
}
dfs2(root);
for(int i=;i<=n;i++)
{
printf("%d\n",max(dp[i],dp1[i]));
}
}
return ;
}
Don't waste your time on a man/woman, who isn't willing to waste their time on you.
不要为那些不愿在你身上花费时间的人而浪费你的时间
--snowy
2019-01-18 07:49:19