POJ 1741 Tree （点分治）

Tree

Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 24238 Accepted: 8053

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

5 4
1 2 3
1 3 1
1 4 2
3 5 1
0 0

Sample Output

　　这是一道点分治（树分治）的经典题目。
　　考虑一条路径
　　如果它完全位于子树B，C，D中，那么这条路径可以在向下层递归时被覆盖到
　　所以我们只需要考虑过根节点A的路径：
　　设deep[x]表示x到root的距离，则我们要求deep[x]+deep[y]<=k&&（x,y不属于同一子树）的点对(x,y)的个数
　　再考虑一下则转化成：[deep[x]+deep[y]<=k]的个数-[deep[x]+deep[y]<=k&&(x,y在同一子树)]的个数
　　这样递归分治求解即可
　　
　　POJ传送门：http://poj.org/problem?id=1741
　

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int inf=1e6;int n,k,root,a,b,vv,tot,cnt,sum,record,ans;int head[10005],nxt[20005],point[20005],weight[20005],dis[10005],deep[10005],size[10005];bool vis[10005];int read(){    int x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}    return x*f;}void clear(){    tot=0;cnt=0;root=0;record=inf;ans=0;    memset(head,0,sizeof(head));    memset(vis,0,sizeof(vis));}void addedge(int x,int y,int value){    tot++;nxt[tot]=head[x];head[x]=tot;point[tot]=y;weight[tot]=value;    tot++;nxt[tot]=head[y];head[y]=tot;point[tot]=x;weight[tot]=value;}void findroot(int now,int father){    size[now]=1;    int maxx=0;    for(int tmp=head[now];tmp;tmp=nxt[tmp]){        int v=point[tmp];        if(!vis[v]&&v!=father){            findroot(v,now);            size[now]+=size[v];            maxx=max(maxx,size[v]);        }    }    maxx=max(maxx,sum-size[now]);    if(maxx<record)record=maxx,root=now;}void getdeep(int now,int father){    deep[++cnt]=dis[now];    for(int tmp=head[now];tmp;tmp=nxt[tmp]){        int v=point[tmp];        if(!vis[v]&&v!=father){            dis[v]=dis[now]+weight[tmp];            getdeep(v,now);        }    }}int query(int now,int value){    cnt=0;dis[now]=value;    getdeep(now,0);    sort(deep+1,deep+cnt+1);    int t=0,l=1,r=cnt;    while(l<r){        if(deep[l]+deep[r]<=k)t+=(r-l),l++;        else r--;    }    return t;}void slove(int now){    vis[now]=true;    ans+=query(now,0);    for(int tmp=head[now];tmp;tmp=nxt[tmp]){        int v=point[tmp];        if(!vis[v]){            ans-=query(v,weight[tmp]);            sum=size[v];            root=0;            record=inf;            findroot(v,root);            slove(root);        }    }}int main(){    while(1){        n=read();k=read();        clear();        if(n==0)break;        for(int i=1;i<n;i++)        {            a=read(),b=read(),vv=read();            addedge(a,b,vv);        }        sum=n;        findroot(1,0);        slove(root);        printf("%d\n",ans);    }}