P2634 [国家集训队]聪聪可可（树上启发式合并）

Posted 2021-09-07 Jozky86

P2634 [国家集训队]聪聪可可（树上启发式合并）

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：

没有修改，统计边长为3的倍数，经典的树上路径统计，树上启发式请求一战
但是调了一阵子没调出来，我对dsu的理解还是不够深，

代码：

待修改代码

#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= 1e5 + 9;
vector<PII> vec[maxn];
int sz[maxn];
int dis[maxn];
int son[maxn];
int Son;
int num[3];
int ans= 0;
void dfs_son(int u, int fa)
{
    sz[u]= 1;

    for (auto it : vec[u]) {
        int v= it.first;
        int w= it.second;
        if (v == fa)
            continue;
        dis[u]= dis[fa] + w;
        dfs_son(v, u);
        sz[u]+= sz[v];
        if (sz[v] > sz[son[u]])
            son[u]= v;
    }
}
void cal(int u, int fa, int LCA)
{
    int x= ((dis[LCA] - dis[u]) % 3 + 3) % 3;
    ans+= num[x];
    for (auto it : vec[u]) {
        int v= it.first;
        if (v != fa && v != son[u])
            cal(v, u, LCA);
    }
}
void add(int u, int fa, int val, int LCA)
{
    int Dis= (dis[u] - dis[LCA]) % 3;
    num[Dis]+= val;
    for (auto it : vec[u]) {
        int v= it.first;
        if (v != fa && v != son[u])
            add(v, u, val, LCA);
    }
}
void work(int u, int fa, int keep)
{
    for (auto it : vec[u]) {
        int v= it.first;
        if (v != fa && v != son[u])
            work(v, u, 0);
    }
    if (son[u]) {
        work(son[u], u, 1);
        Son= son[u];
    }

    add(u, fa, 1, u);
    Son= 0;
    cal(u, fa, u);
    if (!keep) {
        add(u, fa, -1, u);
    }
}
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});
    }
    dis[1]= 1;
    dfs_son(1, 0);
    work(1, 0, 1);
    ll a= ans + n;
    ll b= 1ll * n * n;
    ll g= __gcd(a, b);
    printf("%lld/%lld\\n", a / g, b / g);
    //Time_test();
}

洛谷上的AC代码：

#include <bits/stdc++.h>
using namespace std;
#define For(pos) for (int k= First[pos]; k; k= Next[k])
const int Maxn= 2e4 + 5;
int n, First[Maxn], to[Maxn * 2], Next[Maxn * 2], W[Maxn * 2], cnt;
int son[Maxn], size[Maxn], deep[Maxn];
inline void add(int z, int y, int w)
{
    Next[++cnt]= First[z];
    First[z]= cnt;
    to[cnt]= y;
    W[cnt]= w;
}
int ans= 0, P;
inline int R()
{
    char c;
    int sign= 1, res= 0;
    while ((c= getchar()) > '9' || c < '0')
        if (c == '-')
            sign= -1;
    res+= c - '0';
    while ((c= getchar()) >= '0' && c <= '9')
        res= res * 10 + c - '0';
    return res * sign;
}
void deal(int pos, int father)
{
    size[pos]= 1;
    For(pos)
    {
        if (to[k] == father)
            continue;
        deep[to[k]]= deep[pos] + W[k];
        deal(to[k], pos);
        size[pos]+= size[to[k]];
        if (size[son[pos]] < size[to[k]])
            son[pos]= to[k];
    }
}
int q[4];

inline void cal(int pos, int LCA)
{
    int x= (2 * deep[LCA] - deep[pos]) % 3 + 3;
    x= x % 3;
    ans+= q[x];
}
void work(int pos, int father, bool ca, int LCA)
{
    if (ca)
        cal(pos, LCA);
    else
        q[deep[pos] % 3]++;
    For(pos)
    {
        if (to[k] == father)
            continue;
        work(to[k], pos, ca, LCA);
    }
}
void dfs(int pos, int father, bool heavy)
{
    For(pos) if (to[k] != father && to[k] != son[pos])
    {
        dfs(to[k], pos, 0);
    }
    if (son[pos])
        dfs(son[pos], pos, 1);

    For(pos)
    {
        if (to[k] == father || to[k] == son[pos])
            continue;
        work(to[k], pos, 1, pos);
        work(to[k], pos, 0, pos);
    }
    cal(pos, pos);
    q[deep[pos] % 3]++;
    if (!heavy)
        q[0]= q[1]= q[2]= 0;
}
int main()
{
    n= R();
    int a, b, w;
    for (int i= 1; i < n; i++) {
        a= R();
        b= R();
        w= R();
        add(a, b, w);
        add(b, a, w);
    }
    deal(1, 0);
    dfs(1, 0, 1);
    ans= ans * 2 + n;
    int di= n * n;
    for (int i= 2; i <= ans; i++)
        while (ans % i == 0 && di % i == 0) {
            ans/= i;
            di/= i;
        }
    if (ans == di)
        puts("1/1");
    else
        printf("%d/%d\\n", ans, di);
}