hdu-2169 Computer(樹形dp+樹的直徑)

標籤：

題目連結：

Computer

Time Limit: 1000/1000 MS (Java/Others)   

 Memory Limit: 32768/32768 K (Java/Others)

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 51 12 13 11 1 

 

Sample Output 32344  題意： 給一棵樹的邊的情況,問樹上每個點到其餘點的最大距離是多少;  思路： 樹上一點到其他點的最大距離的這條鏈一定至少有一個端點是直徑的端點,所以,先dfs找到一個端點;再來一遍bfs找到所有點到這個端點的距離,這些距離可能就是最後的答案,也可能不是,然後再從直徑的另一個端點出發,找到所有點到這個端點的距離,和上個距離取最大就是結果了;  AC代碼：

/*    2196    15MS    1976K    2205 B    G++    2014300227*/#include <bits/stdc++.h>using namespace std;const int N=1e4+4;typedef long long ll;const double PI=acos(-1.0);int n,vis[N],head[N],cnt,a,b,dp[N],fp[N],mmax,ending;queue<int>qu;struct Edge{    int to,next,val;};Edge edge[2*N];void add_edge(int s,int e,int va){    edge[cnt].to=e;    edge[cnt].next=head[s];    edge[cnt].val=va;    head[s]=cnt++;}void dfs(int x,int leng){    if(leng>mmax)    {        ending=x;        mmax=leng;    }    vis[x]=1;    for(int i=head[x];i!=-1;i=edge[i].next)    {        int y=edge[i].to;        if(!vis[y])        {            dfs(y,leng+edge[i].val);        }    }}void bfs1(){    qu.push(ending);    dp[ending]=0;    vis[ending]=0;    while(!qu.empty())    {        int fr=qu.front();        qu.pop();        for(int i=head[fr];i!=-1;i=edge[i].next)        {            int y=edge[i].to;            if(vis[y])            {                dp[y]=dp[fr]+edge[i].val;                qu.push(y);                vis[y]=0;            }        }    }}void bfs2(){    int start,mmx=0;    for(int i=1;i<=n;i++)    {        if(dp[i]>mmx)        {            start=i;            mmx=dp[i];        }    }    qu.push(start);    fp[start]=0;    vis[start]=1;    while(!qu.empty())    {        int fr=qu.front();        qu.pop();        for(int i=head[fr];i!=-1;i=edge[i].next)        {            int y=edge[i].to;            if(!vis[y])            {                fp[y]=fp[fr]+edge[i].val;                dp[y]=max(dp[y],fp[y]);                vis[y]=1;                qu.push(y);            }        }    }}int main(){    while(scanf("%d",&n)!=EOF)    {        for(int i=1;i<=n;i++)        {            vis[i]=0;            head[i]=-1;        }        cnt=0;        mmax=0;        for(int i=2;i<=n;i++)        {            scanf("%d%d",&a,&b);            add_edge(i,a,b);            add_edge(a,i,b);        }        dfs(1,0);        bfs1();        bfs2();        for(int i=1;i<=n;i++)        {            printf("%d\n",dp[i]);        }    }    return 0;}

 

  

hdu-2169 Computer(樹形dp+樹的直徑)