聪聪可可

题目传送门

解题思路

点分治。分别统计(各个点到根的距离%3)的值为0，1，2的个数，然后统计不在同一颗子树中余数相加再%3的值为0的个数x，因为两个人可以选同一个点，所以x+n即为路径为3倍数的点对的个数。概率即为(x+n)/(n*n)。

代码如下

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;

inline int read(){
    int res = 0, w = 0; char ch = 0;
    while(!isdigit(ch)){
        w |= ch == '-', ch = getchar();
    }
    while(isdigit(ch)){
        res = (res << 3) + (res << 1) + (ch ^ 48);
        ch = getchar();
    }
    return w ? -res : res;
}

template <typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans % lyd;
}

ll get_inv(ll b, ll mod){
    return fpow(b, mod - 2, mod);
}

const int N = 20005;
struct T{
    int r, w;
    T(int r, int w): r(r), w(w){}
};
vector<T> g[N];
int rt, num, n, sum;
int siz[N];
bool vis[N];

void find_root(int x, int fa)
{
    siz[x] = 1;
    int max_son = 0;
    for(int i = 0; i < g[x].size(); i ++){
        int r = g[x][i].r;
        if(r == fa || vis[r]) continue;
        find_root(r, x);
        siz[x] += siz[r];
        max_son = max(max_son, siz[r]);
    }
    max_son = max(max_son, sum - siz[x]);
    if(max_son < num){
        num = max_son;
        rt = x;
    }
}

int d[N], b[N];
vector<int> vec;

void get_dis(int x, int fa, int s)
{
    for(int i = 0; i < g[x].size(); i ++){
        int r = g[x][i].r;
        if(r == fa || vis[r]) continue;
        d[r] = (d[x] + g[x][i].w) % 3, b[r] = s;
        vec.push_back(r);
        get_dis(r, x, s);
    }
}

int cnt[N][3], ans;
int e[3];

void calc()
{
    memset(cnt, 0, sizeof(cnt));
    memset(e, 0, sizeof(e));
    for(int i = 0; i < vec.size(); i ++){
        cnt[b[vec[i]]][d[vec[i]] % 3] ++;
        e[d[vec[i]] % 3] ++;
    }
    for(int i = 0; i < vec.size(); i ++){
        int t = d[vec[i]] % 3;
        for(int j = 0; j < 3; j ++){
            if((t + j) % 3 == 0)
                ans += e[j] - cnt[b[vec[i]]][j];
        }
    }
}

void dfs(int x)
{
    num = n;
    find_root(x, 0);
    int t = rt;
    vis[t] = true;
    vec.clear();
    vec.push_back(t);
    d[t] = b[t] = 0;
    for(int i = 0; i < g[t].size(); i ++){
        int r = g[t][i].r;
        if(!vis[r]){
            d[r] = g[t][i].w % 3, b[r] = r;
            vec.push_back(r);
            get_dis(r, t, r);
        }
    }
    calc();
    for(int i = 0; i < g[t].size(); i ++){
        int r = g[t][i].r;
        if(!vis[r]){
            sum = siz[r];
            dfs(r);
        }
    }
}

int main()
{
    scanf("%d", &n);
    for(int i = 1; i < n; i ++){
        int x, y, w;
        scanf("%d%d%d", &x, &y, &w);
        g[x].push_back(T(y, w));
        g[y].push_back(T(x, w));
    }
    sum = n;
    dfs(1);
    int b = n * n;
    ans += n;
    int w = __gcd(ans, b);
    printf("%d/%d\n", ans / w, b / w);
    return 0;
}

标签：洛谷,int,max,son,++,vec,聪聪,ans,P2634
来源： https://www.cnblogs.com/whisperlzw/p/11545274.html