hdu5468（容斥原理）-2015 ACM/ICPC Asia Regional Shanghai Online

题意：给一颗以1为根的树，以每一个节点为子树，计算该子树的所有节点value与该节点value互质的节点个数。

假设第i个节点value为vi：

1、由莫比乌斯函数容斥求互质个数，显然需要预处理vi的无平方素数的因子（miu[vi] !=0的因子）；

ans[i] = sigma( miu[d] * count[d] )，d为vi的无平方素数因子，count[d]为以第i个节点为根的子树中value为d的倍数的个数.

2、由上面的公式，容斥的时候需要知道count[d], 所以在遍历当前子树儿子节点之前统计所有的count1[d],在遍历完儿子节点之后统计所有的count2[d]，

显然，count2[d]-count1[d]即为上式中的count[d].

ps:子树中vi自身也算，当且仅当vi=1的时候成立.

参考代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<sstream>
#include<string>
#include<bitset>
using namespace std;

typedef long long LL;

const LL LINF = (1LL <<63);
const int INF = 1 << 31;

const int NS = 100010;
const int MS = 19;
const int MOD = 1000000007;

const int EDGE_MAX = NS;
struct graphEdge
{
int pst;
int next;
};

struct ForwardStart
{
int top;
int head[EDGE_MAX];
graphEdge edge[EDGE_MAX << 1];

void init(int len)
{
top = 0;
memset(head, -1, sizeof(int) * len);
}

void addEdge(int u, int v)
{
edge[top].pst = v;
edge[top].next = head[u];
head[u] = top++;
}

void printAll()
{
printf("top = %d\n", top);
for(int i = 1; i < EDGE_MAX; i++)
{
if(-1 != head[i])
{
printf("head[%2d]'son:%2d", i, edge[head[i]].pst);
for(int j = edge[head[i]].next; j != -1; j = edge[j].next)
{
printf(",%2d", edge[j].pst);
}
puts("");
}
}
}
}cTree;

bitset<NS> isPrime;
vector<int> fac[NS];
int miu[NS];

void prepare()
{
isPrime.set();
isPrime[1] = false;
miu[1] = 1;

for(int i = 2; i < NS; i++)
{
if(isPrime[i])
{
for(int j = i; j < NS; j+=i)
{
isPrime[j] = false;

int k = j / i;
if(k % i)
{
miu[j] = -miu[k];
}
else
{
miu[j] = 0;
}
fac[j].push_back(i);

}
}
else
{
if(miu[i] != 0)
{
for(int j = i; j < NS; j+=i)
{
fac[j].push_back(i);
}
}
}
}
}

int n;
int val[NS];
int ans[NS];
int sz[NS];
int dp[NS];

void dfs(int rt, int fa)
{
vector<int> temp;
sz[rt] = 1;

int value = val[rt];
int len = fac[value].size();
for(int i = 0; i < len; i++)
{
int d = fac[value][i];
int cnt = dp[d];
temp.push_back(cnt);

dp[d] += 1;
}

for(int i = cTree.head[rt]; i != -1; i = cTree.edge[i].next)
{
int cson = cTree.edge[i].pst;
if(cson == fa) continue;

dfs(cson, rt);
sz[rt] += sz[cson];
}

ans[rt] = sz[rt];
for(int i = 0; i < len; i++)
{
int d = fac[value][i];
int cnt = dp[d] - temp[i];
if(cnt > 0)
{
ans[rt] += miu[d] * cnt;
}
}
return ;
}

int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif

prepare();
int nCase = 1;
while(~scanf("%d", &n))
{
cTree.init(n + n + 2);
int u, v;
for(int i = 1; i < n; i++)
{
scanf("%d %d", &u, &v);
cTree.addEdge(u, v);
cTree.addEdge(v, u);
}
for(int i = 1; i <= n; i++)
{
scanf("%d", &val[i]);
if(val[i] < 0)
{
val[i] = - val[i];
}
}

memset(dp, 0, sizeof(dp));
dfs(1, -1);

printf("Case #%d:", nCase++);
for(int i = 1; i <= n; i++)
{
printf(" %d", ans[i]);
}
printf("\n");
}
return 0;
}