# 【BZOJ】1468: Tree（点分治）

http://www.lydsy.com/JudgeOnline/problem.php?id=1468

$$T(n)=aT(n/b)+O(D(n))$$

$$T(n)=aT(n/a)+O(n)$$

#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define print(a) printf("%d", a)
#define dbg(x) cout << (#x) << " = " << (x) << endl
#define error(x) (!(x)?puts("error"):0)
#define rdm(x, i) for(int i=ihead[x]; i; i=e[i].next)
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }

const int N=40005, oo=~0u>>1;
int ihead[N], cnt, K;
struct dat { int next, to, w; }e[N<<1];
void add(int u, int v, int w) {
}

int dep[N], d[N], cdep, ans, mn;
int root, sz[N], vis[N];
void getroot(int x, int fa, int sum) {
sz[x]=1; int y, mx=0;
rdm(x, i) if(!vis[y=e[i].to] && e[i].to!=fa) {
getroot(y, x, sum);
sz[x]+=sz[y];
mx=max(mx, sz[y]);
}
mx=max(mx, sum-mx);
if(mx<mn) mn=mx, root=x;
}
void getdep(int x, int fa) {
dep[++cdep]=d[x]; int y; //printf("x:%d\tfa:%d\tdep:%d\n", x, fa, dep[x]);
rdm(x, i) if(!vis[y=e[i].to] && e[i].to!=fa) {
d[y]=d[x]+e[i].w;
getdep(y, x);
}
}
int cal(int x, int last=0) {
cdep=0; d[x]=last; //printf("root is:%d\n", x);
getdep(x, -1); //puts("==========================");
int ret=0, front=1, tail=cdep;
sort(dep+1, dep+1+cdep);
while(front<tail) {
while(front<tail && dep[tail]+dep[front]>K) --tail;
ret+=tail-front;
++front;
}
return ret;
}
void dfs(int x, int all) {
vis[x]=1; int y;
ans+=cal(x); //printf("root:%d\n", x);
rdm(x, i) if(!vis[y=e[i].to]) {
ans-=cal(y, e[i].w);
int s=sz[y]>sz[x]?all-sz[x]:sz[y];
root=0; mn=oo; getroot(y, x, s);
dfs(root, s);
}
}

int main() {
int n=getint();
rep(i, n-1) { int u=getint(), v=getint(), w=getint(); add(u, v, w); }
getroot((n+1)>>1, -1, n);
dfs(root, n);
printf("%d\n", ans);
return 0;
}

## 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

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

5

## Source

posted @ 2014-12-18 17:39  iwtwiioi  阅读(557)  评论(0编辑  收藏  举报