【BZOJ 2152】 聪聪可可

【题目链接】

https://www.lydsy.com/JudgeOnline/problem.php?id=2152

【算法】

点分治

【代码】

#include<bits/stdc++.h>
using namespace std;
#define MAXN 20010

int i,n,u,v,w,ans1,ans2,g,tot,len,root;
int head[MAXN],size[MAXN],weight[MAXN],d[MAXN],sum[MAXN];
bool visited[MAXN];

struct Edge
{
    int to,w,nxt;
} e[MAXN<<];

inline int gcd(int x,int y)
{
    if (y == ) return x;
    else return gcd(y,x % y);
}
inline void addedge(int u,int v,int w)
{
    tot++;
    e[tot] = (Edge){v,w,head[u]};
    head[u] = tot;
}
inline void getroot(int u,int fa,int total)
{
    int i,v;
    size[u] = ;
    weight[u] = ;
    for (i = head[u]; i; i = e[i].nxt)
    {
        v = e[i].to;
        if (v != fa && !visited[v])
        {
            getroot(v,u,total);
            size[u] += size[v];
            weight[u] = max(weight[u],size[v]);
        }
    }
    weight[u] = max(weight[u],total - size[u]);
    if (weight[u] < weight[root]) root = u;
}
inline void dfs(int u,int fa)
{
    int i,v,w;
    d[++len] = sum[u];
    for (i = head[u]; i; i = e[i].nxt)
    {
        v = e[i].to;
        w = e[i].w;
        if (v != fa && !visited[v])
        {
            sum[v] = (sum[u] + w) % ;
            dfs(v,u);
        }
    }
}
inline int calc(int u)
{
    int i;
    int ret = ;
    int cnt[];
    memset(cnt,,sizeof(cnt));
    len = ;
    dfs(u,);
    for (i = ; i <= len; i++) cnt[d[i]]++;
     for (i = ; i <= len; i++) ret += cnt[( - d[i]) % ];
    return ret;
}
inline void work(int u)
{
    int i,v,w;
    visited[u] = true;
    sum[u] = ;
    ans1 += calc(u);
    for (i = head[u]; i; i = e[i].nxt)
    {
        v = e[i].to;
        w = e[i].w;
        if (!visited[v])
        {
            sum[v] = w % ;
            ans1 -= calc(v);
            root = ;
            getroot(v,,size[v]);
            work(root);
        }
    }
}

int main()
{

    scanf("%d",&n);
    for (i = ; i < n; i++)
    {
        scanf("%d%d%d",&u,&v,&w);
        addedge(u,v,w);
        addedge(v,u,w);
    }
    memset(visited,false,sizeof(visited));
    size[] = weight[] = n;
    getroot(,,n);
    ans1 = ;
    ans2 = n * n;
    work(root);
    g = gcd(ans1,ans2);
    ans1 /= g; ans2 /= g;
    printf("%d/%d\n",ans1,ans2);

    return ;
}