hdu2196Computer（树形dp）

标签：树形dp

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.

5
1 1
2 1
3 1
1 1

3
2
3
4
4

其实就是 递归的应用，不够看别人思路挺新颖的

  http://www.cnblogs.com/kuangbin/archive/2012/08/28/2659915.html

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
typedef __int64 ll;

using namespace std;

#define N 10005

int head[N],num,n;
int big[N],small[N];
int bigid[N],smallid[N];

struct stud{
  int to,len;
  int next;
}e[N*2];

inline void add(int s,int to,int len)
{
	e[num].to=to;
	e[num].len=len;
	e[num].next=head[s];
	head[s]=num++;
}

void dfs1(int x,int father)
{
	  big[x]=small[x]=0;
	  int i;
	  for(i=head[x];i!=-1;i=e[i].next)
	  {
		  int to=e[i].to;
	  	  if(to==father) continue;
	  	  dfs1(to,x);
	  	  if(big[to]+e[i].len>small[x])
		  {
		  	 small[x]=big[to]+e[i].len;
		  	 smallid[x]=to;
		  	 if(small[x]>big[x])
			 {
			 	swap(small[x],big[x]);
			 	swap(smallid[x],bigid[x]);
			 }
		  }
	  }
}

void dfs2(int x,int father)
{
    int i;
    for(i=head[x];i!=-1;i=e[i].next)
	{
		int to=e[i].to;
		if(to==father) continue;

		if(bigid[x]==to)
		{
			if(small[x]+e[i].len>small[to])
			{
				small[to]=small[x]+e[i].len;
				smallid[to]=x;
				if(small[to]>big[to])
				{
					swap(big[to],small[to]);
					swap(bigid[to],smallid[to]);
				}
			}
		}
		else
		{
			if(big[x]+e[i].len>small[to])
			{
				small[to]=big[x]+e[i].len;
				smallid[to]=x;
				if(small[to]>big[to])
				{
					swap(big[to],small[to]);
					swap(bigid[to],smallid[to]);
				}
			}
		}
		dfs2(to,x);
	}
}

int main()
{
	int i,j;
	while(~scanf("%d",&n))
	{
		memset(head,-1,sizeof(head));
		int len,to;
		num=0;
		for(i=2;i<=n;i++)
		{
			scanf("%d%d",&to,&len);
			add(i,to,len);
			add(to,i,len);
		}

		dfs1(1,-1);
		dfs2(1,-1);

		for(i=1;i<=n;i++)
			printf("%d\n",big[i]);
	}
	return 0;
}

hdu2196Computer（树形dp）

标签：树形dp

原文地址：http://blog.csdn.net/u014737310/article/details/43940313