POJ 1741 Tree 树分治

Tree

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://poj.org/problem?id=1741

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.

Under two situations the player could score one point.

⋅1. If you touch a buoy before your opponent, you will get one point. For example if your opponent touch the buoy #2 before you after start, he will score one point. So when you touch the buoy #2, you won't get any point. Meanwhile, you cannot touch buoy #3 or any other buoys before touching the buoy #2.

⋅2. Ignoring the buoys and relying on dogfighting to get point. If you and your opponent meet in the same position, you can try to fight with your opponent to score one point. For the proposal of game balance, two players are not allowed to fight before buoy #2 is touched by anybody.

There are three types of players.

Speeder: As a player specializing in high speed movement, he/she tries to avoid dogfighting while attempting to gain points by touching buoys.
Fighter: As a player specializing in dogfighting, he/she always tries to fight with the opponent to score points. Since a fighter is slower than a speeder, it's difficult for him/her to score points by touching buoys when the opponent is a speeder.
All-Rounder: A balanced player between Fighter and Speeder.

There will be a training match between Asuka (All-Rounder) and Shion (Speeder).
Since the match is only a training match, the rules are simplified: the game will end after the buoy #1 is touched by anybody. Shion is a speed lover, and his strategy is very simple: touch buoy #2,#3,#4,#1 along the shortest path.

Asuka is good at dogfighting, so she will always score one point by dogfighting with Shion, and the opponent will be stunned for T seconds after dogfighting. Since Asuka is slower than Shion, she decides to fight with Shion for only one time during the match. It is also assumed that if Asuka and Shion touch the buoy in the same time, the point will be given to Asuka and Asuka could also fight with Shion at the buoy. We assume that in such scenario, the dogfighting must happen after the buoy is touched by Asuka or Shion.

The speed of Asuka is V1 m/s. The speed of Shion is V2 m/s. Is there any possibility for Asuka to win the match (to have higher score)?

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

HINT

题意

给你一颗树，问你有多少个对点的距离小于k

题解：

树分治，可以去看漆子超的论文，讲的很详细

http://wenku.baidu.com/link?url=7KOPn20aLvKK5PqDmuLjIyj4sqZ6CL1H9qP__JSGvX-AWgX7LR6gC-BZ3PTVCP2ojBHxKZcJ5U3csiRjuspqcoFJfswO7JaEIQyKlxwUzBi

代码

#include<iostream>
#include<stdio.h>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;

#define maxn 10005
struct Edge
{
    int v,w;
    Edge(int vi, int wi) :v(vi), w(wi){}
    Edge(){}
    bool operator <(const Edge &y)const
    {
        return w<y.w;
    }
};
vector<Edge>E[maxn];
int n,k;
int f[maxn];
int sum,root,ans,cnt;
int vis[maxn],son[maxn],d[maxn],deep[maxn];
void init()
{
    root = ans = cnt = 0;
    for(int i=0;i<=n;i++)
        E[i].clear();
    memset(f,0,sizeof(f));
    memset(vis,0,sizeof(vis));
    memset(son,0,sizeof(son));
    memset(d,0,sizeof(d));
    memset(deep,0,sizeof(deep));
}
void getroot(int x,int fa)
{
    son[x]=1;f[x]=0;
    for(int i=0;i<E[x].size();i++)
    {
        int v = E[x][i].v,w = E[x][i].w;
        if(v == fa || vis[v])continue;
        getroot(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 getdeep(int x,int fa)
{
    deep[++deep[0]]=d[x];
    for(int i=0;i<E[x].size();i++)
    {
        int v = E[x][i].v,w = E[x][i].w;
        if(v == fa || vis[v])continue;
        d[v]=d[x]+w;
        getdeep(v,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=1,r=deep[0];
    while(l<r)
    {
        if(deep[l]+deep[r]<=k)
            t+=r-l,l++;
        else r--;
    }
    return t;
}
void solve(int x)
{
    ans += cal(x,0);
    vis[x]=1;
    for(int i=0;i<E[x].size();i++)
    {
        int v = E[x][i].v,w = E[x][i].w;
        if(vis[v])continue;
        ans -= cal(v,w);
        sum = son[v];
        root = 0;
        getroot(v,root);
        solve(root);
    }
}
int main()
{
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        if(n==0&&k==0)break;
        init();
        for(int i=1;i<n;i++)
        {
            int x,y,z;
            scanf("%d%d%d",&x,&y,&z);
            E[x].push_back(Edge(y,z));
            E[y].push_back(Edge(x,z));
        }
        sum=n,f[0]=999999;
        getroot(1,0);
        solve(root);
        printf("%d
",ans);
    }
}