POJ 2196 Computer（搜尋-深度優先搜尋）

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Computer
Problem DescriptionA 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. 
InputInput 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. 
OutputFor each case output N lines. i-th line must contain number Si for i-th computer (1<=i<=N). 
Sample Input

51 12 13 11 1

 
Sample Output

32344

 
Authorscnu 
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題目大意：

告訴你一棵樹，問你某個點最遠能到達多遠？

解題思路：

先從1號點出發，找出各個點到1號點的距離，最遠的那個點必然是樹的主枝幹上的一點。

從樹的主幹上的那點出發，距離最遠的另一點必然是樹主枝乾的另一點。

接下來的答案就是每個點到兩個主幹點的距離取大。

解題代碼：

#include <iostream>#include <cstdio>#include <climits>#include <map>#include <vector>#include <algorithm>using namespace std;const int maxn=11000;struct edge{    int u,v,w;    int next;    edge(int u0=0,int v0=0,int w0=0){ u=u0;v=v0;w=w0;}}e[maxn*2];int n,cnt,head[maxn],d[maxn],dx[maxn],dy[maxn];void initial(){    cnt=0;    for(int i=0;i<=n;i++) head[i]=-1;}void addedge(int u,int v,int w){    e[cnt]=edge(u,v,w);e[cnt].next=head[u];head[u]=cnt++;}void input(){    int x,y,w0;    for(int i=2;i<=n;i++){        scanf("%d%d",&y,&w0);        addedge(i,y,w0);        addedge(y,i,w0);    }}void dfs(int u,int fa,int dis,int *d){    for(int i=head[u];i!=-1;i=e[i].next){        int v=e[i].v,w=e[i].w;        if(v!=fa) dfs(v,u,d[v]=dis+w,d);    }}void solve(){    int x=1,y=1;    dfs(1,-1,d[1]=0,d);    for(int i=1;i<=n;i++) if(d[x]<d[i]) x=i;    dfs(x,-1,dx[x]=0,dx);    for(int i=1;i<=n;i++) if(dx[y]<dx[i]) y=i;    dfs(y,-1,dy[y]=0,dy);    for(int i=1;i<=n;i++) d[i]=max(dx[i],dy[i]);    for(int i=1;i<=n;i++) printf("%d\n",d[i]);}int main(){    while(scanf("%d",&n)!=EOF){        initial();        input();        solve();    }    return 0;}