楼教主男人八题--POJ1741--点分治

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.

裸的点分治。

代码如下：

#include
#include
#include
#include
#define maxn 10006
using namespace std;
int n,K,tot,core,Min,sum,ans,lnk[maxn],nxt[maxn*2],son[maxn*2],w[maxn*2],num[maxn],deep[maxn],d[maxn],f[maxn];
bool vis[maxn];
void add(int x,int y,int z){
    nxt[++tot]=lnk[x];son[tot]=y;w[tot]=z;lnk[x]=tot;
}
void get_core(int x,int fa){
    num[x]=1;f[x]=0;
    for(int j=lnk[x];j;j=nxt[j]) if((son[j]!=fa)&&(vis[son[j]])){
        get_core(son[j],x);
        num[x]+=num[son[j]];f[x]=max(f[x],num[son[j]]);
    }
    f[x]=max(f[x],sum-num[x]);
    if(f[x]void get_deep(int x,int fa){
    deep[++deep[0]]=d[x];
    for(int j=lnk[x];j;j=nxt[j]) if((fa!=son[j])&&(vis[son[j]])){
        d[son[j]]=d[x]+w[j];
        get_deep(son[j],x);
    }
}
int cal(int x,int now){
    d[x]=now;deep[0]=0;int s=0;
    get_deep(x,x);
    sort(deep+1,deep+1+deep[0]);
    int i=1,j=deep[0];
    while(iwhile((iK))j--;
        s+=j-(i++);
    }
    return s;
}
void dfs(int x){
    vis[x]=0;ans+=cal(x,0);
    for(int j=lnk[x];j;j=nxt[j]) if(vis[son[j]]){
        ans-=cal(son[j],w[j]);sum=num[son[j]];
        Min=1e9;get_core(son[j],son[j]);
        dfs(core);
    }
}
int _read(){
    int sum=0;char ch=getchar();
    while(!(ch>='0'&&ch<='9'))ch=getchar();
    while(ch>='0'&&ch<='9')sum=sum*10+ch-48,ch=getchar();
    return sum;
}
int main(){
    freopen("tree_divide.in","r",stdin);
    freopen("tree_divide.out","w",stdout);
    n=_read();K=_read();
    while(n||K){
        tot=0;memset(lnk,0,sizeof(lnk));memset(vis,1,sizeof(vis));
        for(int i=1,x,y,z;i1e9;sum=n;get_core(1,1);ans=0;
        dfs(core);
        printf("%d\n",ans);
        n=_read();K=_read();
    }
    return 0;
}