POJ 1741 Tree

Tree

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 20124		Accepted: 6613

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

Sample Output

Source

LouTiancheng@POJ

IOI论文下载

#include<cstdio>
#include<cstring>
#include<algorithm>
#define m(x) memset(x,0,sizeof x)
using namespace std;
int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9') ch=getchar();
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x;
}
const int N=1e4+10;
struct node{
    int v,w,next;
}e[N<<1];
int n,K,tot,root,sum,ans,head[N],son[N],f[N],d[N],dep[N];
bool vis[N];
void add(int x,int y,int z){
    e[++tot].v=y;e[tot].w=z;e[tot].next=head[x];head[x]=tot;
    e[++tot].v=x;e[tot].w=z;e[tot].next=head[y];head[y]=tot;
} 
void get_root(int x,int fa){//寻找重心 
//重心，就是删掉此结点后，剩下的结点最多的树结点个数最小
    son[x]=1;f[x]=0;
    for(int i=head[x],v;i;i=e[i].next){
        if((v=e[i].v)==fa||vis[v]) continue;
        get_root(v,x);
        son[x]+=son[v];
        f[x]=max(f[x],son[v]); 
    }
    f[x]=max(f[x],sum-son[x]);
    if(f[x]<f[root]) root=x;
}
void get_deep(int x,int fa){
    dep[++dep[0]]=d[x];
    for(int i=head[x],v;i;i=e[i].next){
        if((v=e[i].v)==fa||vis[v]) continue;
        d[v]=d[x]+e[i].w;
        get_deep(v,x);
    }
}
int calc(int x,int now){
    d[x]=now;dep[0]=0;
    get_deep(x,0);
    sort(dep+1,dep+dep[0]+1);
    int t=0,l,r;
    for(l=1,r=dep[0];l<r;){
        if(dep[l]+dep[r]<=K){t+=r-l;l++;}
        else r--;
    }
    return t;
}
void work(int x){//分治 
    ans+=calc(x,0);
    vis[x]=1;
    for(int i=head[x],v;i;i=e[i].next){
        if(vis[v=e[i].v]) continue;
        ans-=calc(v,e[i].w);
        sum=son[v];
        get_root(v,root=0);
        work(root);
    }
}
void Cl(){
    ans=0;tot=0;root=0;
    m(head);m(vis);
}
int main(){
    while(Cl(),n=read(),K=read()){
        for(int i=1,x,y,z;i<n;i++) x=read(),y=read(),z=read(),add(x,y,z);
        sum=n;f[0]=0x7fffffff;
        get_root(1,0);
        work(root);
        printf("%d\n",ans);
    }
    return 0;
}

UPD.2017-05-05

#include<cstdio>
#include<cstring>
#include<algorithm>
#define m(s) memset(s,0,sizeof s);
using namespace std;
template <typename T>
inline void read(T &x){
    register char ch=getchar();x=0;
    while(ch<'0'||ch>'9') ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
}
const int N=1e4+5;
int n,k,ans,sum,root;
int siz[N],f[N],d[N],dep[N];bool vis[N];
struct edge{int v,w,next;}e[N<<1];int tot,head[N];
inline void add(int x,int y,int z){
    e[++tot].v=y;e[tot].w=z;e[tot].next=head[x];head[x]=tot;
    e[++tot].v=x;e[tot].w=z;e[tot].next=head[y];head[y]=tot;
}
void findroot(int x,int fa){
    siz[x]=1;f[x]=0;
    for(int i=head[x];i;i=e[i].next){
        if(e[i].v==fa||vis[e[i].v]) continue;
        findroot(e[i].v,x);
        siz[x]+=siz[e[i].v];
        f[x]=max(f[x],siz[e[i].v]);
    }
    f[x]=max(f[x],sum-siz[x]);
    if(f[x]<f[root]) root=x;
}
void getdep(int x,int fa){
    dep[++dep[0]]=d[x];
    for(int i=head[x];i;i=e[i].next){
        if(e[i].v==fa||vis[e[i].v]) continue;
        d[e[i].v]=d[x]+e[i].w;
        getdep(e[i].v,x);
    }
}
inline int calc(int x,int w){
    d[x]=w;dep[0]=0;getdep(x,0);
    stable_sort(dep+1,dep+dep[0]+1);
    int res=0;
    for(int l=1,r=dep[0];l<r;) if(dep[l]+dep[r]<=k) res+=r-l++;else r--;
    return res;
}
void solve(int x){
    ans+=calc(x,0);
    vis[x]=1;
    for(int i=head[x];i;i=e[i].next){
        if(vis[e[i].v]) continue;
        ans-=calc(e[i].v,e[i].w);
        f[root=0]=sum=siz[e[i].v];
        findroot(e[i].v,0);
        solve(root);
    }
}
inline void clr(){
    tot=0;m(head);m(vis);
}
int main(){
    for(read(n),read(k);n&&k;read(n),read(k),clr()){
        for(int i=1,x,y,z;i<n;i++) read(x),read(y),read(z),add(x,y,z);
        f[root=0]=sum=n;findroot(1,0);
        ans=0;solve(root);
        printf("%d\n",ans);
    }
    return 0;
}

__EOF__

本文作者：shenben
本文链接：https://www.cnblogs.com/shenben/p/6269606.html
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