hdu 2196 Computer 树形DP 求每个点的最远点距离 两次dfs的经典树形dp

Computer

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4945    Accepted Submission(s): 2481

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

 

Author

scnu

 

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题意：给出一棵边带权的树，求出每个点的最远点距离。

最新代码：2016/10/25

#include<cstdio>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>using namespace std;#define for0(a, n) for (int (a) = 0; (a) < (n); (a)++)#define for1(a, n) for (int (a) = 1; (a) <= (n); (a)++)#define mes(a,x,s)  memset(a,x,(s)*sizeof a[0])#define mem(a,x)  memset(a,x,sizeof a)#define ysk(x)  (1<<(x))const int maxn=10000    ;int N,fir[maxn+5];int nedge,nex[2*maxn+10],dp_far[maxn+10][2];struct Edge{    int to,w;}e[2*maxn+10];void add_edge(int x,int y,int w){    e[nedge].to=y;    e[nedge].w=w;    nex[nedge]=fir[x];    fir[x]=nedge++;}void insert(int val,int &x0,int &x1){    if(val>x0)    {        x1=x0;        x0=val;    }    else  if(val>x1)    {        x1=val;    }}void dfs(int x,int fa){    dp_far[x][0]=dp_far[x][1]=0;    for(int i=fir[x];~i;i=nex[i])    {        int y=e[i].to;if(y==fa)  continue;        int w=e[i].w;        dfs(y,x);        insert(dp_far[y][0]+w,dp_far[x][0],dp_far[x][1]);    }}void dfs2(int x,int fa){    for(int i=fir[x];~i;i=nex[i])    {        int y=e[i].to;if(y==fa)  continue;        int w=e[i].w;        if(dp_far[y][0]+w==dp_far[x][0]) insert(dp_far[x][1]+w,dp_far[y][0],dp_far[y][1]);        else   insert(dp_far[x][0]+w,dp_far[y][0],dp_far[y][1]);        dfs2(y,x);    }}int main(){   std::ios::sync_with_stdio(false);   int x,w;   while(cin>>N)   {       nedge=0;mes(fir,-1,N+1);       for(int i=2;i<=N;i++)       {           cin>>x>>w;           add_edge(i,x,w);           add_edge(x,i,w);       }       dfs(1,-1);       dfs2(1,-1);       for1(i,N)  printf("%d\n",dp_far[i][0]);   }   return 0;}

老代码：

#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>#include<climits>#include<queue>#include<vector>#include<map>#include<sstream>#include<set>#include<stack>#include<cctype>#include<utility>#pragma comment(linker, "/STACK:102400000,102400000")#define PI (4.0*atan(1.0))#define eps 1e-10#define sqr(x) ((x)*(x))#define FOR0(i,n)  for(int i=0 ;i<(n) ;i++)#define FOR1(i,n)  for(int i=1 ;i<=(n) ;i++)#define FORD(i,n)  for(int i=(n) ;i>=0 ;i--)#define  lson   ind<<1,le,mid#define rson    ind<<1|1,mid+1,ri#define MID   int mid=(le+ri)>>1#define zero(x)((x>0? x:-x)<1e-15)#define mk    make_pair#define _f     first#define _s     second#define ysk(x)  (1<<(x))using namespace std;//const int INF=    ;typedef long long ll;//const ll inf =1000000000000000;//1e15;//ifstream fin("input.txt");//ofstream fout("output.txt");//fin.close();//fout.close();//freopen("a.in","r",stdin);//freopen("a.out","w",stdout);const int INF =0x3f3f3f3f;const int maxn=10000+20    ;//const int maxm=    ;struct Edge{    int from,to,w;    Edge(){}    Edge(int from,int to,int w):from(from),to(to),w(w){}};vector<Edge>edges;vector<int >G[maxn];inline void add_edge(int s,int t,int w){    edges.push_back(Edge(s,t,w));    edges.push_back(Edge(t,s,w));    int m=edges.size();    G[s].push_back(m-2);    G[t].push_back(m-1);}int n,dp[maxn];struct  DPD{    int a[2];    int p[2];    void clear()    {        a[0]=a[1]=0;        p[0]=p[1]=-1;    }    void insert(int num,int x)    {        if(num>=a[0])        {            a[1]=a[0];            p[1]=p[0];            a[0]=num;            p[0]=x;        }        else if(a[0]>num&&num>=a[1])        {            a[1]=num;            p[1]=x;        }    }}  dpd[maxn] ;int dfs(int x,int fa){    dpd[x].clear();    for(int i=0;i<G[x].size();i++)    {        Edge & e=edges[G[x][i]];        int & y=e.to;        int & w=e.w;        if(y==fa)  continue;        dpd[x].insert(dfs(y,x)+w,y  );    }    return dpd[x].a[0];}void DP(int x,int fa,int dis){     dp[x]=max(dpd[x].a[0], dis  );   DPD ret=dpd[x];     ret.insert(dis,fa);    for(int i=0;i<G[x].size();i++)    {        Edge & e=edges[G[x][i]];        int & y=e.to;        int & w=e.w;        if(y==fa)  continue;        if( y!=ret.p[0] )  DP(y,x,ret.a[0]+w  );        else     DP(y,x,ret.a[1]+w);    }}int main(){  while(~scanf("%d",&n))  {      for(int i=1;i<=n;i++)  G[i].clear();      edges.clear();      for(int i=2;i<=n;i++)      {          int x,w;          scanf("%d%d",&x,&w);          add_edge(i,x,w);      }      dfs(1,-1);      DP(1,-1,0);      for(int i=1;i<=n;i++)      {          printf("%d\n",dp[i]);      }  }    return 0;}