Tree
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 20098 | Accepted: 6608 |
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.
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.
The last test case is followed by two zeros.
Output
For each test case output the answer on a single line.
题意：给一颗带权树，求树上长度不超过L的路径条数
首先有一个有树高限制时的树形DP做法..........
对于一条树路径 只有经过或不经过一个点的情况
考虑经过一个点的路径，可以由其他点到它的两条路径拼出来
对于不经过的情况 把一棵树按这个点拆成好几棵分治
每次对于当前子树选择树的重心，最多递归logn次，而每层最多只有n个点(每层的所有子树组成整棵树)，复杂度O(logn*处理每层的复杂度)
过程：
1.求重心
2.处理经过当前点的路径
3.对子树分治
每次分治的各个子树是互不影响的，vis[i]表示i这个点已经分治过了
注意：既然你的写法是先找重心在递归，那么一定要rt=0;dfsRt(v,0);dfsSol(rt);是rt啊啊啊啊啊不是v了
对于本题，处理经过点u的路径时，先dfs子树中所有点对深度，排序两个指针往里扫计算<=L的，在减去在同一颗子树里的(同样计算)
总复杂度O(nlog^2n)
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int N=10005,INF=1e9+5; inline int read(){ char c=getchar();int x=0,f=1; while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();} while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();} return x*f; } int n,L,u,v,w; struct edge{ int v,w,ne; }e[N<<1]; int h[N],cnt; inline void ins(int u,int v,int w){ cnt++; e[cnt].v=v;e[cnt].w=w;e[cnt].ne=h[u];h[u]=cnt; cnt++; e[cnt].v=u;e[cnt].w=w;e[cnt].ne=h[v];h[v]=cnt; } int size[N],d[N],vis[N],root,sum; void dfsRoot(int u,int fa){ size[u]=1;d[u]=0; for(int i=h[u];i;i=e[i].ne){ int v=e[i].v; if(vis[v]||v==fa) continue; dfsRoot(v,u); size[u]+=size[v]; d[u]=max(d[u],size[v]); } d[u]=max(d[u],sum-size[u]); if(d[u]<d[root]) root=u; } int deep[N],a[N]; void dfsDeep(int u,int fa){ a[++a[0]]=deep[u]; for(int i=h[u];i;i=e[i].ne){ int v=e[i].v; if(vis[v]||v==fa) continue; deep[v]=deep[u]+e[i].w; dfsDeep(v,u); } } int cal(int u,int now){ deep[u]=now;a[0]=0; dfsDeep(u,0); sort(a+1,a+1+a[0]); int l=1,r=a[0],ans=0; while(l<r){ if(a[l]+a[r]<=L) ans+=r-l,l++; else r--; } return ans; } int ans; void dfsSol(int u){//printf("dfs %d ",u); vis[u]=1; ans+=cal(u,0); for(int i=h[u];i;i=e[i].ne){ int v=e[i].v; if(vis[v]) continue; ans-=cal(v,e[i].w); sum=size[v]; root=0;dfsRoot(v,0); dfsSol(root); } } int main(){ //freopen("in.txt","r",stdin); while(true){ n=read();L=read();if(n==0) break; cnt=0;memset(h,0,sizeof(h)); memset(vis,0,sizeof(vis)); ans=0; for(int i=1;i<=n-1;i++) u=read(),v=read(),w=read(),ins(u,v,w); sum=n; root=0;d[0]=INF; dfsRoot(1,0); dfsSol(root); printf("%d ",ans); } }
还有一种做法，用treap维护，考虑经过每个点的路径时，建一颗treap维护长度，每个点加上之前遍历过的这点的子树中<L-deep[v]+1的，最后再把这棵子树的所有深度加入treap
// // main.cpp // treap // // Created by Candy on 2017/1/9. // Copyright ? 2017年 Candy. All rights reserved. // #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> using namespace std; #define lc t[x].l #define rc t[x].r const int N=1e5+5,INF=1e9; int read(){ char c=getchar();int x=0,f=1; while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();} while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();} return x*f; } struct node{ int l,r,v,w,size,rnd; }t[N]; int sz,root; inline void update(int x){t[x].size=t[lc].size+t[rc].size+t[x].w;} inline void lturn(int &x){ int c=rc;rc=t[c].l;t[c].l=x; t[c].size=t[x].size;update(x);x=c; } inline void rturn(int &x){ int c=lc;lc=t[c].r;t[c].r=x; t[c].size=t[x].size;update(x);x=c; } void ins(int &x,int v){ if(x==0){ x=++sz; t[x].l=t[x].r=0; t[x].v=v;t[x].w=t[x].size=1; t[x].rnd=rand(); return; } t[x].size++; if(v==t[x].v) t[x].w++; else if(v<t[x].v){ ins(lc,v); if(t[lc].rnd<t[x].rnd) rturn(x); }else{ ins(rc,v); if(t[rc].rnd<t[x].rnd) lturn(x); } } int que(int x,int v){//cnt of <v if(!x) return 0; if(t[x].v==v) return t[lc].size; if(v<t[x].v) return que(lc,v); else return t[lc].size+t[x].w+que(rc,v); } int n,L,u,v,w; struct edge{ int v,w,ne; }e[N<<1]; int h[N],cnt; inline void ins(int u,int v,int w){ cnt++; e[cnt].v=v;e[cnt].w=w;e[cnt].ne=h[u];h[u]=cnt; cnt++; e[cnt].v=u;e[cnt].w=w;e[cnt].ne=h[v];h[v]=cnt; } int vis[N],size[N],f[N],sum,rt; void dfsRoot(int u,int fa){ size[u]=1;f[u]=0; for(int i=h[u];i;i=e[i].ne){ int v=e[i].v; if(vis[v]||v==fa) continue; dfsRoot(v,u); size[u]+=size[v]; f[u]=max(f[u],size[v]); } f[u]=max(f[u],sum-size[u]); if(f[u]<f[rt]) rt=u; } int ans,deep[N]; void dfsDeep(int u,int fa,int p){ if(p==0) ans+=que(root,L-deep[u]+1); else ins(root,deep[u]); for(int i=h[u];i;i=e[i].ne){ int v=e[i].v; if(vis[v]||v==fa) continue; deep[v]=deep[u]+e[i].w; dfsDeep(v,u,p); } } void dfsSol(int u){//printf("sol %d ",u); vis[u]=1; sz=root=0; ins(root,0); for(int i=h[u];i;i=e[i].ne){ int v=e[i].v; if(vis[v]) continue; deep[v]=e[i].w; dfsDeep(v,u,0); dfsDeep(v,u,1); } for(int i=h[u];i;i=e[i].ne){ int v=e[i].v; if(vis[v]) continue; sum=size[v]; rt=0;dfsRoot(v,u); dfsSol(rt); } } int main(){ //freopen("in.txt","r",stdin); while(true){ n=read();L=read();if(n==0) break; cnt=0;memset(h,0,sizeof(h)); memset(vis,0,sizeof(vis)); ans=0; for(int i=1;i<=n-1;i++) u=read(),v=read(),w=read(),ins(u,v,w); sum=n; rt=0;f[0]=INF; dfsRoot(1,0); dfsSol(rt); printf("%d ",ans); } }