【BZOJ-1468】Tree 树分治

1468: Tree

Time Limit: 10 Sec  Memory Limit: 64 MB
Submit: 1025  Solved: 534
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Description

给你一棵TREE,以及这棵树上边的距离.问有多少对点它们两者间的距离小于等于K

Input

N(n<=40000) 接下来n-1行边描述管道,按照题目中写的输入 接下来是k

Output

一行,有多少对点之间的距离小于等于k

Sample Input

7
1 6 13
6 3 9
3 5 7
4 1 3
2 4 20
4 7 2
10

Sample Output

5

HINT

Source

LTC男人八题系列

Solution

写LTC还不如直接写ACRush,楼教主的男人八题.暑假看过什么都看不懂,现在似乎可以做了?

裸的树分治,统计一下路径的权值和,排序计算答案即可

Code

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
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;
}
#define maxn 80010
int n,root,siz,ans,stack[maxn],top,K;
struct Edgenode{int to,next,val;}edge[maxn<<1];
int head[maxn],cnt=1;
void add(int u,int v,int w)
{cnt++; edge[cnt].val=w; edge[cnt].to=v; edge[cnt].next=head[u]; head[u]=cnt;}
void insert(int u,int v,int w)
{add(u,v,w);add(v,u,w);}
int gcd(int a,int b)
{if (b==0) return a; return gcd(b,a%b);}
int size[maxn],maxx[maxn],val[maxn]; bool visit[maxn];
void DFSroot(int now,int last)
{
    size[now]=1; maxx[now]=0;
    for (int i=head[now]; i; i=edge[i].next)
        if (edge[i].to!=last && !visit[edge[i].to])
            {
                DFSroot(edge[i].to,now);
                size[now]+=size[edge[i].to];
                maxx[now]=max(maxx[now],size[edge[i].to]);
            }
    maxx[now]=max(maxx[now],siz-size[now]);
    if (maxx[now]<maxx[root]) root=now;
}
void DFSval(int now,int last)
{
    stack[++top]=val[now];
    for (int i=head[now]; i; i=edge[i].next)
        if (edge[i].to!=last && !visit[edge[i].to])
            {    
                val[edge[i].to]=val[now]+edge[i].val;
                DFSval(edge[i].to,now);
            }
}
int Getans(int now,int va)
{
    val[now]=va; top=0; DFSval(now,0);
    sort(stack+1,stack+top+1);
    int re=0,l=1,r=top;
    while (l<r) if (stack[l]+stack[r]<=K) re+=r-l,l++; else r--;
    return re;
}
void work(int now)
{
    ans+=Getans(now,0);
    visit[now]=1;
    for (int i=head[now]; i; i=edge[i].next)
        if (!visit[edge[i].to])
            {
                ans-=Getans(edge[i].to,edge[i].val);
                root=0; siz=size[edge[i].to];
                DFSroot(edge[i].to,0); work(root);
            }
}
int main()
{
    n=read();
    for (int u,v,w,i=1; i<=n-1; i++) 
        u=read(),v=read(),w=read(),insert(u,v,w);
    K=read();
    maxx[0]=siz=n;
    DFSroot(1,0);
    work(root);
    printf("%d\n",ans);
    return 0;
}

吃饭前想10分钟Rush出来...手速不够TAT

——It's a lonely path. Don't make it any lonelier than it has to be.
posted @ 2016-05-08 18:46  DaD3zZ  阅读(213)  评论(0编辑  收藏  举报