Poj1741[Tree]题解--点分治||Treap

【链接】
poj1741

【题目大意】
给你一棵树，对于这课树有n个顶点，定义dist(u,v)= 节点u和v之间的最小距离。再给你一个数k，使dist(u,v)<=k为一个有效对数，计算对于给定树有效的对数

【解题报告】
此题就是Treap的启发式合并和点分治的裸题，不用多说贴代码。

Treap:

#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>using namespace std;const int maxn=10005,maxm=20005;int n,m,tot,ans,lnk[maxn],w[maxm],son[maxm],nxt[maxm];struct Treap{    Treap* son[2];    int x,s,w,ran;    int Cmp(int k) {if (k<x) return 0; if (k>x) return 1; return -1;}    void Updata() {s=son[0]->s+son[1]->s+w;}}a[maxn*6],*Null=a,*ro[maxn];void Add(int x,int y,int z){    w[++tot]=z; son[tot]=y; nxt[tot]=lnk[x]; lnk[x]=tot;}inline int Read(){    int res=0;    char ch=getchar();    while (ch<'0'||ch>'9') ch=getchar();    while (ch>='0'&&ch<='9') res=res*10+ch-48,ch=getchar();    return res;}void Turn(Treap* &k,int d){    Treap* t=k->son[d]; k->son[d]=t->son[d^1]; t->son[d^1]=k;    t->s=k->s; k->Updata(); k=t;}void New(Treap* &k,int x,int p){    ++tot; k=&a[tot]; k->s=k->w=p; k->x=x; k->ran=rand(); k->son[0]=k->son[1]=Null;}void Insert(Treap* &k,int x,int p){    if (k==Null) {New(k,x,p); return;}    int d=k->Cmp(x); k->s+=p;    if (d<0) k->w+=p;     else      {        Insert(k->son[d],x,p);        if (k->son[d]->ran<k->ran) Turn(k,d);      }}int Ask(Treap* k,int x){    if (k==Null) return 0;    int d=k->Cmp(x);    if (d<0) return k->son[0]->s+k->w;     else if (d) return k->son[0]->s+k->w+Ask(k->son[1],x);    return Ask(k->son[0],x);}void Join(Treap* &k1,Treap* k2){    if (k2==Null) return;    Insert(k1,k2->x,k2->w);    Join(k1,k2->son[0]); Join(k1,k2->son[1]);}int Count(Treap* k1,Treap* k2,int x){    if (k2==Null) return 0;    return k2->w*Ask(k1,x-k2->x)+Count(k1,k2->son[0],x)+Count(k1,k2->son[1],x);}void Dfs(int x,int dep,int fa){    for (int j=lnk[x]; j; j=nxt[j])     if (son[j]!=fa)      {        Dfs(son[j],dep+w[j],x);        if (ro[x]->s<ro[son[j]]->s) swap(ro[x],ro[son[j]]);        ans+=Count(ro[x],ro[son[j]],2*dep+m); Join(ro[x],ro[son[j]]);      }    ans+=Ask(ro[x],dep+m);    Insert(ro[x],dep,1);}void Work(){    memset(lnk,0,sizeof(lnk)); tot=0;    for (int i=1; i<=n; i++) ro[i]=Null;    for (int i=1,x,y,z; i<n; i++) {x=Read();y=Read();z=Read(); Add(x,y,z); Add(y,x,z);}    tot=ans=0; Dfs(1,0,0);    printf("%d\n",ans);    n=Read(); m=Read();}int main(){    n=Read(); m=Read();    while (n||m) Work();    return 0;}

点分治：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn=10005,maxm=20005,INF=((1<<30)-1)*2+1;int n,m,tot,rot,ans,nn,p[maxn],f[maxn],f_son[maxn],dep[maxn],lnk[maxn],son[maxm],nxt[maxm],w[maxm];bool vis[maxn];inline 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;}void add_(int x,int y,int z){    w[++tot]=z; son[tot]=y; nxt[tot]=lnk[x]; lnk[x]=tot;}void getrot(int x,int fa){    f_son[x]=1; f[x]=0;    for (int j=lnk[x]; j; j=nxt[j])     if (!vis[son[j]]&&son[j]!=fa)      {        getrot(son[j],x);        f_son[x]+=f_son[son[j]];        f[x]=max(f[x],f_son[son[j]]);      }    f[x]=max(f[x],nn-f_son[x]);    if (f[x]<f[rot]) rot=x;}void getdep(int x,int fa){    dep[++dep[0]]=p[x];    for (int j=lnk[x]; j; j=nxt[j])     if (!vis[son[j]]&&son[j]!=fa)      {        p[son[j]]=p[x]+w[j];        getdep(son[j],x);      }}int cal(int x,int v){    p[x]=v; dep[0]=0;    getdep(x,0);    sort(dep+1,dep+1+dep[0]);    int L=1,R=dep[0],sum=0;    while (L<R)     if (dep[L]+dep[R]<=m) {sum+=R-L; L++;}  else R--;    return sum;}void solve(int x){    vis[x]=1; ans+=cal(x,0);    for (int j=lnk[x]; j; j=nxt[j])     if (!vis[son[j]])      {        ans-=cal(son[j],w[j]);        nn=f_son[son[j]]; rot=0;        getrot(son[j],0);        solve(rot);      }}void work_(){    memset(vis,0,sizeof(vis));    memset(lnk,0,sizeof(lnk));    tot=0;    for (int i=1; i<n; i++)    {        int x=read_(),y=read_(),z=read_();        add_(x,y,z); add_(y,x,z);    }    nn=n; f[0]=INF; rot=ans=0;    getrot(1,0);    solve(rot);    printf("%d\n",ans);    n=read_(); m=read_();}int main(){    n=read_(); m=read_();    while (n||m) work_();    return 0;}