BZOJ2599: [IOI2011]Race

n<=200000的边权树上问长度=K<=1000000的链中边数最少的。

点分治。开个桶统计各个长度的答案。

#include<string.h>
#include<stdlib.h>
#include<stdio.h>
#include<math.h>
#include<algorithm>
//#include<bitset>
//#include<queue>
//#include<iostream>
using namespace std;

int n,K;
#define maxn 200011
#define maxk 1000011
struct Edge{int to,v,next;}edge[maxn<<1]; int first[maxn],le=2;
void in(int x,int y,int v) {Edge &e=edge[le]; e.to=y; e.v=v; e.next=first[x]; first[x]=le++;}
void insert(int x,int y,int v) {in(x,y,v); in(y,x,v);}

int size[maxn]; bool die[maxn];
void getsize(int x,int fa)
{
    size[x]=1;
    for (int i=first[x];i;i=edge[i].next)
    {
        const Edge &e=edge[i]; if (e.to==fa || die[e.to]) continue;
        getsize(e.to,x); size[x]+=size[e.to];
    }
}

int getroot(int x,int fa,int tot)
{
    for (int i=first[x];i;i=edge[i].next)
    {
        const Edge &e=edge[i]; if (e.to==fa || die[e.to]) continue;
        if (size[e.to]*2>tot) return getroot(e.to,x,tot);
    }
    return x;
}

int f[maxk];
void calc(int x,int fa,int now,int cur)
{
    if (now>K) return;
    f[K]=min(f[K],f[K-now]+cur);
    for (int i=first[x];i;i=edge[i].next)
    {
        const Edge &e=edge[i]; if (e.to==fa || die[e.to]) continue;
        calc(e.to,x,now+e.v,cur+1);
    }
}

void gao(int x,int fa,int now,int v,bool op)
{
    if (now>K) return;
    if (op) f[now]=min(f[now],v);
    else (now!=K && (f[now]=0x3f3f3f3f));
    for (int i=first[x];i;i=edge[i].next)
    {
        const Edge &e=edge[i]; if (e.to==fa || die[e.to]) continue;
        gao(e.to,x,now+e.v,v+1,op);
    }
}

void cd(int x)
{
    die[x]=1; f[0]=0;
    for (int i=first[x];i;i=edge[i].next)
    {
        const Edge &e=edge[i]; if (die[e.to]) continue;
        calc(e.to,x,e.v,1); gao(e.to,x,e.v,1,1);
    }
    for (int i=first[x];i;i=edge[i].next)
    {
        const Edge &e=edge[i]; if (die[e.to]) continue;
        gao(e.to,x,e.v,0,0);
    }
    for (int i=first[x];i;i=edge[i].next)
    {
        const Edge &e=edge[i]; if (die[e.to]) continue;
        getsize(e.to,x); cd(getroot(e.to,x,size[e.to]));
    }
}

int main()
{
    scanf("%d%d",&n,&K);
    for (int i=1,x,y,v;i<n;i++)
    {
        scanf("%d%d%d",&x,&y,&v); x++; y++;
        insert(x,y,v);
    }
    getsize(1,0);
    for (int i=1;i<=K;i++) f[i]=0x3f3f3f3f;
    cd(getroot(1,0,size[1]));
    printf("%d\n",f[K]==0x3f3f3f3f?-1:f[K]);
    return 0;
}