HDU5468：DFS序+容斥原理

最新推荐文章于 2019-08-13 14:41:01 发布

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探讨PuzzledElena算法挑战，涉及图论、质因数分解与互质数计数，通过深度优先搜索（DFS）优化计算树状结构中节点间互质关系。

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Puzzled Elena

Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2103 Accepted Submission(s): 731

Problem Description

Since both Stefan and Damon fell in love with Elena, and it was really difficult for her to choose. Bonnie, her best friend, suggested her to throw a question to them, and she would choose the one who can solve it.

Suppose there is a tree with n vertices and n - 1 edges, and there is a value at each vertex. The root is vertex 1. Then for each vertex, could you tell me how many vertices of its subtree can be said to be co-prime with itself?
NOTES: Two vertices are said to be co-prime if their values' GCD (greatest common divisor) equals 1.

Input

There are multiply tests (no more than 8).
For each test, the first line has a number n (1≤n≤105), after that has n−1 lines, each line has two numbers a and b (1≤a,b≤n), representing that vertex a is connect with vertex b. Then the next line has n numbers, the ith number indicates the value of the ith vertex. Values of vertices are not less than 1 and not more than 105.

Output

For each test, at first, please output "Case #k: ", k is the number of test. Then, please output one line with n numbers (separated by spaces), representing the answer of each vertex.

Sample Input

Sample Output

Case #1: 1 1 0 0 0

Source

2015 ACM/ICPC Asia Regional Shanghai Online

题解

可以先参考HDU4135：求解区间[a,b]中与k互质的数的个数
做这一题之前可以先解决一个问题：给定任意一个序列，求解k与这个序列互质的数的个数。

举个?：求8，9，12总与3互质的数的个数

我们先统计出质因数

8有质因数2，所以p[2]++；

9有质因数3，所以p[3]++；

12有质因数2，3，所以p[2]++,p[3]++;

所以与3不互质的数的个数为p[3] = 2个（9和12各贡献出了一个3，所以他们与3都不互质）

所以与3互质的数的个数为3 - 2 = 1;

每个质因数可以理解为贡献了一个数

这一题也类似，先DFS序一遍。具体代码参考如下

#include <cstring>
#include <cstdio>
#include <vector>
using namespace std;
typedef long long ll;
int const N = 100000 + 10;
int ne[2*N],to[2*N],first[N],w[N],ans[N],cnt[N];
int tot,n;
vector<int>v[N];
void add(int x,int y){
	ne[++tot] = first[x];
	to[tot] = y;
	first[x] = tot;
}
void Pre(){
	for(int k=1;k<N;k++){
		int n = k;
		for(int i=2;(ll)i*i<=n;i++){
			if(n % i == 0){
				v[k].push_back(i);
				while(n % i == 0)	n /= i;
			}
		}
		if(n > 1)	v[k].push_back(n);
	}
}
int cal(int u,int k){
	int num = v[u].size();  //u的质因数的个数
	int ans = 0;
	for(int i=1;i<(1<<num);i++){
		int mul = 1,bit = 0;
		for(int j=0;j<num;j++)
			if(i&(1<<j)){
				mul *= v[u][j];
				bit++;
			}
		if(bit&1)	ans += cnt[mul];
		else 		ans -= cnt[mul];
		cnt[mul] += k;
	}
	return ans;
}
int DFS(int u,int fa){
	int pre = cal(w[u],0);
	int s = 0;
	for(int i=first[u];i;i=ne[i]){
		int v = to[i];
		if(v == fa)	continue;
		s += DFS(v,u);
	}
	int last = cal(w[u],1);
	ans[u] = s - (last - pre);   //u子树中结点的个数 - 与u不是互质的结点的个数
	if(w[u] == 1)	ans[u]++; //1与本身互质           ==(包括字树与前面的与u不互质的结点的个数-前面与u不互质的结点的个数)
	return s+1;   //返回结点个数（包括自己）
}
int main(){
	int caser = 0;
	Pre();
	while(~scanf("%d",&n)){
		memset(cnt,0,sizeof(cnt));
		tot = 0;
		memset(first,0,sizeof(first));
		for(int i=1;i<=n-1;i++){
			int x,y;
			scanf("%d%d",&x,&y);
			add(x,y);	add(y,x);
		}
		for(int i=1;i<=n;i++)	scanf("%d",&w[i]);
		DFS(1,0);
		printf("Case #%d:",++caser);
		for(int i=1;i<=n;i++)	printf(" %d",ans[i]);
		printf("\n");
	}
	return 0;
}