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 LOCALstartTime= clock();freopen("in.txt", "r", stdin);
#endif
}
void Time_test()
{
#ifdef LOCALendTime= 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);elseq[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");elseprintf("%d/%d\n", ans, di);
}