标签:ons string prime prim max 获得 std 时间复杂度 oid
https://cn.vjudge.net/problem/Gym-101840H
给出一棵树,问有多少对点,将他们之间的边权相乘之后所获得的值仅有两个不同的质因子。
当时没想着用并查集做,写了个点分治= =。我们对于重心的子树挨个搜索,用number数组记录下质因子数量为0, 1, 2的点的数量,对于质因子数量为1的查询时还要知道多少与你当前想要链接的点为相同的质因子,所以用has【i】记录下质因子为 i 的点的数量。那么能与质因子数量为1的点相连的点就是number【1】-has【arr【j】】(arr【j】是当前选择的点所拥有的质因子)。
#include <iostream> #include <cstring> #include <string> #include <algorithm> #include <cmath> #include <cstdio> #include <queue> #include <stack> #include <map> #include <bitset> #define ull unsigned long long #define met(a, b) memset(a, b, sizeof(a)) #define lowbit(x) (x&(-x)) #define MID (l + r) / 2 #define ll long long using namespace std; const int maxn = 1e5 + 7; const ll mod = 1e6 + 3; const int inf = 0x3f3f3f3f; const ll INF = 0x3f3f3f3f3f3f3f3f; struct Edge { int to, dist, net; }edge[maxn * 2]; int n, k; int head[maxn], cnt; int val[maxn]; int dis[maxn], arr[maxn], tail, all, sz[maxn], msz[maxn], rt; bool vis[maxn]; int res; bitset<maxn> is_prime; int prime[maxn], ans; int number[3]; int has[maxn]; int N[40]; void num() { is_prime[1] = 1; for(int i = 2; i < maxn; i++) { if(!is_prime[i]) prime[ans++] = i; for(int j = 0; j < ans && i * prime[j] <= n; j++) { is_prime[i * prime[j]] = 1; if(i % prime[j] == 0) break; } } } void init() { res = 0; for(int i = 0; i <= n; i++) { head[i] = -1; vis[i] = 0; } met(has, 0); cnt = 0; } void addedge(int u, int v, int c) { edge[cnt] = (Edge){v, c, head[u]}; head[u] = cnt++; } void Findrt(int pos, int pre) { sz[pos] = 1; msz[pos] = 0; for(int i = head[pos]; i != -1; i = edge[i].net) { int to = edge[i].to; if(to == pre || vis[to]) continue; Findrt(to, pos); sz[pos] += sz[to]; msz[pos] = max(msz[pos], sz[to]); } msz[pos] = max(msz[pos], all - sz[pos]); if(msz[pos] < msz[rt] || rt == 0) rt = pos; } int calc(int v) { int sum = 0; for(int i = 0; i < ans && sum < 3 && prime[i] <= v; i++) { int flag = 1; while(sum < 3 && v % prime[i] == 0) { if(flag) { flag = 0; N[++sum] = prime[i], v /= prime[i]; } else sum = 3; } } return sum; } void dfs(int pos, int pre, int v) { dis[pos] = dis[pre] + calc(v); if(dis[pos] == 2) arr[++tail] = -1; else if(dis[pos] == 1) arr[++tail] = N[1]; else if(dis[pos] == 0) arr[++tail] = 0; for(int i = head[pos]; i != -1; i = edge[i].net) { int to = edge[i].to; if(vis[to] || to == pre) continue; dfs(to, pos, edge[i].dist); } } void divide(int pos) { vis[pos] = 1; for(int i = head[pos]; i != -1; i = edge[i].net) { int to = edge[i].to; if(vis[to]) continue; tail = 0; dfs(to, pos, edge[i].dist); for(int j = 1; j <= tail; j++) { if(arr[j] == -1) res += number[0] + 1; else if(arr[j] == 0) res += number[2]; else res += number[1] - has[arr[j]]; } for(int j = 1; j <= tail; j++) { if(arr[j] == -1) number[2]++; else if(arr[j] == 0) number[0]++; else number[1]++, has[arr[j]]++; } }
//原本这里是要递归进子树清除对进入下一层的影响,memset时间复杂度太高。但当时调不出来qwq tail = 0; met(has, 0); met(dis, 0); for(int i = 0; i < 3; i++) number[i] = 0; for(int i = head[pos]; i != -1; i = edge[i].net) { int to = edge[i].to; if(vis[to]) continue; all = sz[to]; rt = 0; Findrt(to, 0); divide(rt); } } int main() { freopen("evaluations.in", "r", stdin); num(); int T, k = 0; cin >> T; while(T--) { cin >> n; init(); for(int i = 1; i < n; i++) { int u, v, c; cin >> u >> v >> c; addedge(u, v, c); addedge(v, u, c); } all = n; rt = 0; Findrt(1, 0); divide(rt); printf("Case %d: %d\n", ++k, res); } return 0; }
Half Nice Years Gym - 101840H (点分治 or 并查集)
标签:ons string prime prim max 获得 std 时间复杂度 oid
原文地址:https://www.cnblogs.com/Stay-Online/p/11360332.html