Poj 1741 Tree 点分治 解题报告

Description

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

思路

其实我基本就不会点分治。。

发现确实比链分治简单不少，所谓点分治 对于一条树路径 只有经过或不经过一个点的情况。

对于不经过的情况 把一棵树按这个点拆成好几棵分治就行了，考虑经过这个点的情况，对于这题 可以对这个点延伸出的几棵子树各做一次dfs，记录子树中出现的距离值。
对于一棵树的距离值数组，把它排序求一次ans1，再对每棵子树分别求一个自己对自己的ans2，ans1−∑ans2ans1−∑ans2即为最后的ans

代码

借鉴黄学长，hzwer那么有名我就不贴网址了。

#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>#include<cmath>#include<queue>using namespace std;const int INF=0x7fffffff;const int N=10000+5;using namespace std;int n,k,num,sum,ans,root;int head[N],deep[N],d[N],f[N],son[N],vis[N];struct edge{    int u,next,v;};edge ed[N*3];void build(int u,int v,int w){    ed[++num].u=v;    ed[num].next=head[u];    head[u]=num;    ed[num].v=w;}void insert(int u,int v,int w){    build(u,v,w);    build(v,u,w);}void getroot(int x,int fa){    son[x]=1;f[x]=0;    for (int i=head[x];i;i=ed[i].next)    {        if (ed[i].u==fa||vis[ed[i].u]) continue;        getroot(ed[i].u,x);        son[x]+=son[ed[i].u];        f[x]=max(f[x],son[ed[i].u]);    }    f[x]=max(f[x],sum-son[x]);    if (f[x]<f[root]) root=x;}void getdeep(int x,int fa){    deep[++deep[0]]=d[x];    for (int i=head[x];i;i=ed[i].next)    {        if (ed[i].u==fa||vis[ed[i].u]) continue;        d[ed[i].u]=d[x]+ed[i].v;        getdeep(ed[i].u,x);    }}int cal(int x,int now){    d[x]=now;deep[0]=0;    getdeep(x,0);    sort(deep+1,deep+deep[0]+1);    int t=0,l,r;    for (l=1,r=deep[0];l<r;)    {        if (deep[l]+deep[r]<=k){t+=r-l;l++;}        else r--;    }    return t;}void work(int x){    ans+=cal(x,0);    vis[x]=1;    for (int i=head[x];i;i=ed[i].next)    {        if (vis[ed[i].u]) continue;        ans-=cal(ed[i].u,ed[i].v);        sum=son[ed[i].u];        root=0;        getroot(ed[i].u,root);        work(root);    }}int main(){    while(true)    {        ans=0;root=0;num=0;        memset(vis,0,sizeof(vis));        memset(head,0,sizeof(head));        scanf("%d%d",&n,&k);        if (n==0) break;        for (int i=1;i<n;i++)        {            int u,v,w;            scanf("%d%d%d",&u,&v,&w);            insert(u,v,w);        }        sum=n;f[0]=INF;        getroot(1,0);        work(root);        printf("%d\n",ans);    }    return 0;}