The 2019 ICPC China Nanchang National Invitational and International Silk-Road Programming Contest E. Interesting Trip 长链剖分

 题库链接

考虑莫比乌斯， 套上去之后就是变成了统计长度为d的一共有多少路径， 直接长链剖分， 

在计蒜客上极度卡常， 卡了一万年才卡过去， 现场好像还有用点分治过去的， 这都能过？？

#include
#define LL long long
using namespace std;

const int N = (int)5e5 + 7;
const int M = 30000;

int n, d, a[N], vis[N], miu[M + 1];
int now_val, now_op, now_cnt, now_col;
int len[N], son[N], dp[N], *id, *f[N];
bool ok[N];
LL ans;

vector<int> V[M + 1];
vector<int> P[M + 1];

int edge_tot, head[N];
struct Edge {
    int to, nex;
} e[N << 1];

inline void addEdge(int u, int v) {
    e[edge_tot].to = v;
    e[edge_tot].nex = head[u];
    head[u] = edge_tot++;
}

bool checkNum(int x) {
    for(int i = 2; i * i <= x; i++) {
        if(x % (i * i) == 0) return false;
    }
    return true;
}

void prepare() {
    ok[1] = true;
    for(int i = 2; i <= M; i++) {
        if(checkNum(i)) ok[i] = true;
    }
    for(int i = 1; i <= M; i++) {
        for(int j = 1; j * j <= i; j++) {
            if(i % j) continue;
            if(ok[j]) V[i].push_back(j);
            if(j * j != i && ok[i / j]) V[i].push_back(i / j);
        }
    }
    miu[1] = 1;
    for(int i = 1; i <= M; i++) {
        for(int j = i + i; j <= M; j += i) {
            miu[j] -= miu[i];
        }
    }
}

void gao(int u, int fa) {
    now_cnt++;
    vis[u] = now_col;
    son[u] = len[u] = 0;
    for(int j = head[u], v; ~j; j = e[j].nex) {
        v = e[j].to;
        if(v == fa || a[v] % now_val) continue;
        gao(v, u);
        if(len[v] > len[u]) {
            len[u] = len[v];
            son[u] = v;
        }
    }
    len[u]++;
}

void dfs(int u, int fa) {
    f[u][0] = 1;
    if(son[u]) {
        f[son[u]] = f[u] + 1;
        dfs(son[u], u);
    }
    if(d < len[u]) {
        ans += now_op * f[u][d];
    }
    for(int j = head[u], v; ~j; j = e[j].nex) {
        v = e[j].to;
        if(v == fa || v == son[u] || a[v] % now_val) continue;
        f[v] = id; id += len[v];
        dfs(v, u);
        for(int i = 1; i <= len[v] && i <= d; i++) {
            if(d - i < len[u]) ans += 1LL * now_op * f[v][i - 1] * f[u][d - i];
        }
        for(int i = 1; i <= len[v] && i <= d; i++) {
            f[u][i] += f[v][i - 1];
        }
    }
}

void init() {
    ans = edge_tot = 0;
    for(int i = 1; i <= M; i++) {
        P[i].clear();
    }
    for(int i = 1; i <= n; i++) {
        head[i] = -1;
        vis[i] = 0;
    }
}

int main() {
    prepare();
    int cas = 0;
    int T;
    scanf("%d", &T);
    while(T--) {
        scanf("%d%d", &n, &d);
        init();
        for(int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
            for(auto &t : V[a[i]]) {
                P[t].push_back(i);
            }
        }
        for(int i = 1; i < n; i++) {
            int u, v;
            scanf("%d%d", &u, &v);
            addEdge(u, v);
            addEdge(v, u);
        }
        for(int i = 2; i <= M; i++) {
            if(!P[i].size()) continue;
            now_val = i; now_op = -miu[i]; now_col = i;
            for(auto &Rt : P[i]) {
                if(vis[Rt] == i) continue;
                now_cnt = 0;
                gao(Rt, 0);
                for(int i = 0; i <= now_cnt; i++) dp[i] = 0;
                id = dp;
                f[Rt] = id; id += len[Rt];
                dfs(Rt, 0);
            }
        }
        printf("Case #%d: %lld\n", ++cas, 2 * ans);
    }
    return 0;
}

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