Tree
Give a tree with n vertices,each edge has a length(positive integer less than 1001).
Define dist(u,v)=The min distance between node u and v.
Give an integer k,for every pair (u,v) of vertices is called valid if and only if dist(u,v) not exceed k.
Write a program that will count how many pairs which are valid for a given tree.
Input
The input contains several test cases. The first line of each test case contains two integers n, k. (n<=10000) The following n-1 lines each contains three integers u,v,l, which means there is an edge between node u and v of length l.
The last test case is followed by two zeros.
Output
For each test case output the answer on a single line.
Sample Input
5 4 1 2 3 1 3 1 1 4 2 3 5 1 0 0
Sample Output
8
點分治主要是處理統計樹上路徑這類的問題,對一個無根樹,選擇一點爲根,然後求出所有點到根的距離,然後統計經過根的所有路徑對答案的貢獻,然後刪掉這個點,再選另一個點爲根,反覆直到樹上沒有點。每次選擇的點都是當前樹的重心。具體可以見:點分治詳解
題目鏈接:http://poj.org/problem?id=1741
題目大意:給一棵樹,n個節點,n-1條邊,求樹上距離小於等於k的點對的數量
思路:點分治模板題
dep[i] 爲點 i 到根的距離,把dep從小到大排序後,二分,如果dep[l]+dep[r]<=k,則有r-l個點對符合,累加完後,要再減去所有在同一子樹下的,因爲同一子樹下的點對路徑不經過根。
代碼:
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int N=1e4+10;
const int inf=1e9+10;
int n,k,root,ans,sum,tot,first[N];
int siz[N],f[N],vis[N],d[N],dep[N];
struct node
{
int v,w,nex;
}e[N<<1];
void add(int u,int v,int w)
{
e[tot].v=v,e[tot].w=w;
e[tot].nex=first[u];
first[u]=tot++;
}
void init()
{
f[0]=inf;
tot=ans=root=0;
memset(first,-1,sizeof(first));
memset(vis,0,sizeof(vis));
}
void getroot(int u,int fa) //求重心
{
siz[u]=1,f[u]=0; //siz[i]記錄以i爲根的樹的節點數,f[i]記錄以i爲根的最大子樹的大小
for(int i=first[u];~i;i=e[i].nex)
{
int v=e[i].v;
if(v==fa||vis[v]) continue;
getroot(v,u);
siz[u]+=siz[v];
f[u]=max(f[u],siz[v]);
}
f[u]=max(f[u],sum-siz[u]);
if(f[u]<f[root]) root=u; //把子樹最大值最小的更新爲重心,作爲根
}
void getdep(int u,int fa)
{
dep[++dep[0]]=d[u]; //dep[i]記錄點i到根的距離
for(int i=first[u];~i;i=e[i].nex)
{
int v=e[i].v;
if(v==fa||vis[v]) continue;
d[v]=d[u]+e[i].w;
getdep(v,u);
}
}
int cal(int u,int w) //計算點對數
{
d[u]=w;
dep[0]=0;
getdep(u,0);
sort(dep+1,dep+1+dep[0]);
int l=1,r=dep[0],res=0;
while(l<r)
{
if(dep[l]+dep[r]<=k)
{
res+=r-l;
l++;
}
else r--;
}
return res;
}
void solve(int u)
{
ans+=cal(u,0);
vis[u]=1; //刪除根節點
for(int i=first[u];~i;i=e[i].nex)
{
int v=e[i].v;
if(!vis[v])
{
ans-=cal(v,e[i].w); //減去同一子樹下的
root=0,sum=siz[v];
getroot(v,0);
solve(root);
}
}
}
int main()
{
while(~scanf("%d%d",&n,&k)&&(n+k))
{
init();
int u,v,w;
for(int i=1;i<n;i++)
{
scanf("%d%d%d",&u,&v,&w);
add(u,v,w);
add(v,u,w);
}
sum=n;
getroot(1,0);
solve(root);
printf("%d\n",ans);
}
return 0;
}