P2634 [国家集训队]聪聪可可(点分治做法)

Posted 2021-09-07 Jozky86

P2634 [国家集训队]聪聪可可

题意：

一颗n个点的树，问其中两点之间的边上数的和加起来是3的倍数的点对有多少个？
输出这样的点对所占比例

题解：

因为是求三的倍数，我们num来记录%3=0，1，2的数量，对于u得到的d[]，累加num[3-d[]]的数量。
其实就是常规的点分治模板

代码：

#include <bits/stdc++.h>
#include <unordered_map>
#define debug(a, b) printf("%s = %d\\n", a, b);
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
clock_t startTime, endTime;
//Fe~Jozky
const ll INF_ll= 1e18;
const int INF_int= 0x3f3f3f3f;
void read(){};
template <typename _Tp, typename... _Tps> void read(_Tp& x, _Tps&... Ar)
{
    x= 0;
    char c= getchar();
    bool flag= 0;
    while (c < '0' || c > '9')
        flag|= (c == '-'), c= getchar();
    while (c >= '0' && c <= '9')
        x= (x << 3) + (x << 1) + (c ^ 48), c= getchar();
    if (flag)
        x= -x;
    read(Ar...);
}
template <typename T> inline void write(T x)
{
    if (x < 0) {
        x= ~(x - 1);
        putchar('-');
    }
    if (x > 9)
        write(x / 10);
    putchar(x % 10 + '0');
}
void rd_test()
{
#ifdef LOCAL
    startTime= clock();
    freopen("in.txt", "r", stdin);
#endif
}
void Time_test()
{
#ifdef LOCAL
    endTime= clock();
    printf("\\nRun Time:%lfs\\n", (double)(endTime - startTime) / CLOCKS_PER_SEC);
#endif
}
const int maxn= 2e5 + 9;
vector<PII> vec[maxn];
int vis[maxn], d[maxn], sz[maxn];
int rt, cnt;
int num[4];
void dfs_rt(int u, int fa, int tot)
{
    sz[u]= 1;
    int v, w, n= 0;
    for (auto it : vec[u]) {
        v= it.first;
        if (v == fa || vis[v])
            continue;
        dfs_rt(v, u, tot);
        sz[u]+= sz[v];
        n= max(n, sz[v]);
    }
    n= max(n, tot - sz[u]);
    if (2 * n <= tot)
        rt= u;
}
void dfs_dis(int u, int fa, int dis)
{
    d[++cnt]= dis;
    for (auto it : vec[u]) {
        int v= it.first;
        int w= it.second;
        if (v == fa || vis[v])
            continue;
        dfs_dis(v, u, dis + w);
    }
}
int work(int u, int fa, int tot)
{
    dfs_rt(u, fa, tot);
    u= rt;
    vis[u]= 1;
    ll ans= 0;
    for (auto it : vec[u]) {
        int v= it.first;
        int w= it.second;
        if (vis[v])
            continue;
        cnt= 0;
        dfs_dis(v, u, w);
        sz[v]= cnt;
        for (int i= 1; i <= cnt; i++) {
            ans+= num[(3 - d[i] % 3) % 3];
            if (d[i] % 3 == 0)
                ans++;
        }
        for (int i= 1; i <= cnt; i++) {
            num[d[i] % 3]++;
        }
    }
    num[0]= num[1]= num[2]= 0;
    for (auto it : vec[u]) {
        int v= it.first;
        if (vis[v])
            continue;
        ans+= work(v, u, sz[v]);
    }
    return ans;
}
int main()
{
    //rd_test();
    int n;
    read(n);
    for (int i= 1; i < n; i++) {
        int u, v, w;
        read(u, v, w);
        vec[u].push_back({v, w});
        vec[v].push_back({u, w});
    }
    ll a= 2ll * work(1, 0, n) + n;
    ll b= 1ll * n * n;
    ll g= __gcd(a, b);
    printf("%lld/%lld\\n", a / g, b / g);
    return 0;
    //Time_test();
}