BZOJ3522&4543: [POI2014]Hotel加强版

题解见http://blog.youkuaiyun.com/neither_nor/article/details/51278993

不知道这是什么狗题，时间复杂度貌似是O(N)

#include<iostream>
#include<cstdio>
#include<cstring>
#define M 3000010
#define N 200010
using namespace std;
long long wa[M];
long long *f[N],*g[N],*p=wa; 
long long to[N],nxt[N],pre[N],cnt;
void ae(long long ff,long long tt)
{
    cnt++;
    to[cnt]=tt;
    nxt[cnt]=pre[ff];
    pre[ff]=cnt;
}
long long d[N],fa[N],mx[N],yz[N];
void build(long long x)
{
    long long i,j;
    yz[x]=x;
    for(i=pre[x];i;i=nxt[i])
    {
        j=to[i];
        if(j==fa[x]) continue;
        fa[j]=x;
        build(j);
        if(d[j]>d[mx[x]])
        {
            yz[x]=yz[j];
            mx[x]=j;
        }
    }
    if(mx[x]==x) d[x]=0;
    else d[x]=d[mx[x]]+1;
    for(i=pre[x];i;i=nxt[i])
    {
        j=to[i];
        if(j==fa[x]||(j==mx[x]&&x!=1)) continue;
        p+=(d[j]+5);
        f[yz[j]]=p;p+=5;
        g[yz[j]]=p;
        p+=2*d[j]+5;
    }
}
long long ans;
void dfs(long long x)
{
    long long i,j,k;
    for(i=pre[x];i;i=nxt[i])
    {
        j=to[i];
        if(j==fa[x]) continue;
        dfs(j);
        if(j==mx[x])
        {
            f[x]=f[j]-1;
            g[x]=g[j]+1;
        }
    }
    ans+=g[x][0];f[x][0]=1;
    for(i=pre[x];i;i=nxt[i])
    {
        j=to[i];
        if(j==fa[x]||j==mx[x]) continue;
        ans+=g[x][1];
        for(k=1;k<=d[j];k++)
        {
            ans+=g[j][k]*f[x][k-1];
            ans+=f[j][k]*g[x][k+1];
        }
        for(k=0;k<=d[j];k++)
        {
            g[x][k-1]+=g[j][k];
            g[x][k+1]+=f[j][k]*f[x][k+1];
            f[x][k+1]+=f[j][k];
        }
    }
}
long long n;
int main()
{
    scanf("%lld",&n);
    long long i,j,x,y;
    for(i=1;i<n;i++)
    {
        scanf("%lld%lld",&x,&y);
        ae(x,y);ae(y,x);
    }
    build(1);
    if(n==1) {puts("0");return 0;}
    dfs(1);
    printf("%lld\n",ans);
}