BZOJ4543 POI2014 Hotel加强版 【长链剖分】【DP】*

BZOJ4543 POI2014 Hotel加强版

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Description

同OJ3522
数据范围：n<=100000

Sample Input

7
1 2
5 7
2 5
2 3
5 6
4 5

Sample Output

5

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#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define N 100010
LL pool[N<<4];
LL* top=pool;
LL* get(int len){LL* t=top;top+=len;return t;}
LL *f[N],*g[N];
int n;
LL ans=0;
vector<int> p[N];
int dep[N],lson[N];
void dfs1(int u,int fa){
    dep[u]=0;lson[u]=0;
    for(int i=0;i<p[u].size();i++){
        int v=p[u][i];
        if(v==fa)continue;
        dfs1(v,u);
        dep[u]=max(dep[u],dep[v]+1);
        if(dep[v]>dep[lson[u]])lson[u]=v;
    }
}
void dfs2(int u,int fa,int& maxlen,int blank){
    maxlen=max(maxlen,dep[u]);
    if(lson[u]){
        dfs2(lson[u],u,maxlen,blank+1);
        ans+=g[lson[u]][1];
        f[u]=f[lson[u]]-1;
        f[u][0]=1;
        g[u]=g[lson[u]]+1;
    }else{
        f[u]=get(maxlen+5+blank)+blank;
        g[u]=get(maxlen+5+blank);
        f[u][0]=1;
    }
    for(int i=0;i<p[u].size();i++){
        int v=p[u][i],mxlen=0;
        if(v==fa||v==lson[u])continue;
        dfs2(v,u,mxlen,0);
        for(int j=0;j<dep[v];j++)ans+=f[u][j]*g[v][j+1];
        for(int j=1;j<=dep[v]+1;j++)ans+=g[u][j]*f[v][j-1];
        for(int j=1;j<=dep[v]+1;j++)g[u][j]+=f[u][j]*f[v][j-1];
        for(int j=0;j<=dep[v];j++)f[u][j+1]+=f[v][j];
        for(int j=1;j<=dep[v];j++)g[u][j-1]+=g[v][j];
    }
}
int main(){
    scanf("%d",&n);
    for(int i=1;i<n;i++){
        int u,v;
        scanf("%d%d",&u,&v);
        p[u].push_back(v);
        p[v].push_back(u);
    }
    int mxlen=0;
    dfs1(1,0);
    dfs2(1,0,mxlen,0);
    printf("%lld",ans);
    return 0;
}