hdu 5468 Puzzled Elena(前缀性质+dfs序+容斥)

题目链接：hdu 5468 Puzzled Elena

解题思路

预处理出每个数的因子（注意只需要质因子幂数最大为1的数，例如 6=21∗31 ）然后用一个数组维护，fac[i]表示说当前有fac[i]个数包含i这个因子。

然后dfs遍历这棵树，第一次遍历到节点u时，通过fac数组计算出现在有多少个数与W[u]不互质，待遍历完u的子树后，再计算一次，两个相减即为u子树中与W[u]不互质的数的个数，在处理完该节点后，同时要用W[u]更新fac数组。注意u节点也是u子树中的节点，特殊情况为W[u]=1的时候。

通过fac数组计算不互质数的个数只要通过容斥即可。

代码

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

using namespace std;
const int maxn = 1e5 + 5;
typedef long long ll;

int N, W[maxn], fac[maxn], L[maxn], R[maxn], ans[maxn];
int E, first[maxn], jump[maxn<<2], linker[maxn<<2];
vector<int> G[maxn];

void addEdge(int u, int v) {
    jump[E] = first[u];
    linker[E] = v;
    first[u] = E++;
}

void init () {
    E = 0;
    memset(first, -1, sizeof(first));

    int u, v;
    for (int i = 1; i < N; i++) {
        scanf("%d%d", &u, &v);
        addEdge(u, v);
        addEdge(v, u);
    }
    for (int i = 1; i <= N; i++) scanf("%d", &W[i]);
}

int bitcount(int x) { return x ? bitcount(x>>1) + (x&1) : 0; }

int solve (int x, int a) {
    int n = G[x].size(), ret = 0;
    for (int i = 1; i < (1<<n); i++) {
        int t = 1;
        for (int j = 0; j < n; j++) if (i&(1<<j))
            t *= G[x][j];

        if (bitcount(i)&1) ret += fac[t];
        else ret -= fac[t];
        fac[t] += a;
    }
    return ret;
}

int dfs (int u, int f) {
    int cnt = 0;
    L[u] = solve(W[u], 0);

    for (int i = first[u]; i + 1; i = jump[i]) {
        int v = linker[i];
        if (v == f) continue;
        cnt += dfs(v, u);
    }

    R[u] = solve(W[u], 1);
    ans[u] = cnt - (R[u]-L[u]);
    if (W[u] == 1) ans[u]++;
    return cnt + 1;
}

int main () {
    //for (int i = 1; i < maxn; i++) G[i].clear();
    for (int i = 2; i < maxn; i++) {
        if (G[i].size()) continue;
        for (int j = i; j < maxn; j += i)
            G[j].push_back(i);
    }

    int cas = 1;
    while (scanf("%d", &N) == 1) {
        init();

        memset(fac, 0, sizeof(fac));
        dfs(1, 0);
        printf("Case #%d:", cas++);
        for (int i = 1; i <= N; i++)
            printf(" %d", ans[i]);
        printf("\n");
    }
    return 0;
}