poj 1741 树的分治

思路：这题我是看 漆子超《分治算法在树的路径问题中的应用》写的。

附代码：

#include<iostream>

#include<cstring>

#include<algorithm>

#include<cstdio>

#include<cmath>

#define Maxn 10010

#define Maxm 20010

#define inf 0x7fffffff

using namespace std;

int head[Maxn],vi[Maxn],e,ans,num,k,n;

int mx[Maxn],mi,dis[Maxn],root,size[Maxn];

struct Edge{

    int u,v,val,next;

}edge[Maxm];

void init()

{

    memset(vi,0,sizeof(vi));

    memset(head,-1,sizeof(head));

    memset(mx,0,sizeof(mx));

    memset(dis,0,sizeof(dis));

    e=ans=0;

}

void add(int u,int v,int val)

{

    edge[e].u=u,edge[e].v=v,edge[e].val=val,edge[e].next=head[u],head[u]=e++;

    edge[e].u=v,edge[e].v=u,edge[e].val=val,edge[e].next=head[v],head[v]=e++;

}

void dfssize(int u,int fa)

{

    int i,v;

    size[u]=1;

    mx[u]=0;

    for(i=head[u];i!=-1;i=edge[i].next)

    {

        v=edge[i].v;

        if(v!=fa&&!vi[v])

        {

            dfssize(v,u);

            size[u]+=size[v];

            if(size[v]>mx[u]) mx[u]=size[v];

        }

    }

}

void dfsroot(int r,int u,int fa)

{

    int v,i;

    if(size[r]-size[u]>mx[u]) mx[u]=size[r]-size[u];

    if(mx[u]<mi) mi=mx[u],root=u;

    for(i=head[u];i!=-1;i=edge[i].next)

    {

        v=edge[i].v;

        if(v!=fa&&!vi[v])

        {

            dfsroot(r,v,u);

        }

    }

}

void dfsdis(int u,int d,int fa)

{

    int i,v;

    dis[++num]=d;

    for(i=head[u];i!=-1;i=edge[i].next)

    {

        v=edge[i].v;

        if(v!=fa&&!vi[v])

        {

            dfsdis(v,d+edge[i].val,u);

        }

    }

}

int calc(int u,int d)

{

    int i,j,ret=0;

    num=0;

    dfsdis(u,d,0);

    sort(dis+1,dis+1+num);

    i=1;j=num;

    while(i<j)//单调求点对

    {

        while(dis[i]+dis[j]>k&&i<j)

            j--;

        ret+=j-i;

        i++;

    }

    return ret;

}

void dfs(int u)

{

    int i,v;

    mi=n;

    dfssize(u,0);

    dfsroot(u,u,0);

    ans+=calc(root,0);

    vi[root]=1;

    for(i=head[root];i!=-1;i=edge[i].next)

    {

        v=edge[i].v;

        if(!vi[v])

        {

            ans-=calc(v,edge[i].val);

            dfs(v);

        }

    }

}

int main()

{

    int i,j,u,v,val;

    while(scanf("%d%d",&n,&k)!=EOF,n||k)

    {

        init();

        for(i=1;i<n;i++)

        {

            scanf("%d%d%d",&u,&v,&val);

            add(u,v,val);

        }

        dfs(1);

        printf("%d\n",ans);

    }

    return 0;

}