F - Prime Independence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 128000/64000 KB (Java/Others)
Submit Status
Problem Description
A set of integers is called prime independent if none of its member is a prime multiple of another member. An integer a is said to be a prime multiple of b if,
a = b x k (where k is a prime [1])
So, 6 is a prime multiple of 2, but 8 is not. And for example, {2, 8, 17} is prime independent but {2, 8, 16} or {3, 6} are not.
Now, given a set of distinct positive integers, calculate the largest prime independent subset.
Input
Input starts with an integer T (≤ 20), denoting the number of test cases.
Each case starts with an integer N (1 ≤ N ≤ 40000) denoting the size of the set. Next line contains N integers separated by a single space. Each of these N integers are distinct and between 1 and 500000 inclusive.
Output
For each case, print the case number and the size of the largest prime independent subset.
Sample Input
3 5 2 4 8 16 32 5 2 3 4 6 9 3 1 2 3
Sample Output
Case 1: 3 Case 2: 3 Case 3: 2
Hint
An integer is said to be a prime if it‘s divisible by exactly two distinct integers. First few prime numbers are 2, 3, 5, 7, 11, 13, ...
Dataset is huge, use faster I/O methods.
1 #include <stdio.h>
2 #include <queue>
3 #include <cstring>
4 #include <cstdlib>
5 #include <algorithm>
6 #include <vector>
7 #include <map>
8 #include <set>
9 #include <ctime>
10 #include <cmath>
11 #include <cctype>
12 #include <iostream>
13 using namespace std;
14 typedef long long LL;
15 const int N=5*1e5+10;
16 const int INF=0x3f3f3f3f;
17 const int maxn=40010;
18 int cas=1,T;
19 int id[N];
20 struct maxMacth
21 {
22 int n;
23 int vm[N],um[N];
24 bool vis[N];
25 vector<int>g[N];
26 int dx[N],dy[N],dis;
27 void init(int num)
28 {
29 n=num;
30 memset(vm,-1,sizeof(vm));
31 memset(um,-1,sizeof(um));
32 for(int i=0;i<=n;i++)
33 g[i].clear();
34 }
35 void inserts(int u, int v)
36 {
37 g[u].push_back(v);
38 }
39 bool searchP()
40 {
41 queue<int>q;
42 dis=INF;
43 memset(dx,-1,sizeof(dx));
44 memset(dy,-1,sizeof(dy));
45 for(int i=1;i<=n;i++)
46 if(um[i]==-1)
47 {
48 q.push(i);
49 dx[i]=0;
50 }
51 while(!q.empty())
52 {
53 int u=q.front();q.pop();
54 if(dx[u]>dis) break;
55 for(int i=0;i<g[u].size();i++)
56 {
57 int v = g[u][i];
58 if(dy[v]==-1)
59 {
60 dy[v]=dx[u]+1;
61 if(vm[v]==-1) dis=dy[v];
62 else
63 {
64 dx[vm[v]]=dy[v]+1;
65 q.push(vm[v]);
66 }
67 }
68 }
69 }
70 return dis!=INF;
71 }
72 bool dfs(int u)
73 {
74 for(int i=0;i<g[u].size();i++)
75 {
76 int v = g[u][i];
77 if(!vis[v]&&dy[v]==dx[u]+1)
78 {
79 vis[v]=1;
80 if(vm[v]!=-1&&dy[v]==dis) continue;
81 if(vm[v]==-1||dfs(vm[v]))
82 {
83 vm[v]=u;um[u]=v;
84 return 1;
85 }
86 }
87 }
88 return 0;
89 }
90 int maxMatch()
91 {
92 int res=0;
93 while(searchP())
94 {
95 memset(vis,0,sizeof(vis));
96 for(int i=1;i<=n;i++)
97 if(um[i]==-1&&dfs(i)) res++;
98 }
99 return res;
100 }
101 }MM;
102 int pn,p[N],vis[N],e[N];
103 void init()
104 {
105 memset(vis,0,sizeof(vis));
106 memset(e,0,sizeof(e));
107 pn=0;
108 for(int i=2;i<N;i++) if(!vis[i])
109 {
110 p[pn++]=i;
111 for(int j=i;j<N;j+=i)
112 {
113 vis[j]=1;
114 int x=j;
115 while(x%i==0) x/=i,e[j]++;
116 }
117 }
118 }
119 int main()
120 {
121 // freopen("1.in","r",stdin);
122 // freopen("1.out","w",stdout);
123 init();
124 scanf("%d",&T);
125 while(T--)
126 {
127 int n;
128 scanf("%d",&n);
129 memset(id,0,sizeof(id));
130 for(int i=1,x;i<=n;i++) scanf("%d",&x),id[x]=i;
131 MM.init(n);
132 for(int i=1;i<N;i++) if(id[i])
133 {
134 for(int j=0;i*p[j]<N && j<pn;j++) if(id[i*p[j]])
135 {
136 if(e[i]&1)
137 MM.g[id[i]].push_back(id[i*p[j]]);
138 else
139 MM.g[id[i*p[j]]].push_back(id[i]);
140 }
141 }
142 printf("Case %d: %d\n",cas++,n-MM.maxMatch());
143 // printf("Case %d: %d\n",cas++,n-MM.maxMatch()/2);
144 // cerr<<clock()<<"ms\n";
145 }
146 return 0;
147 }
solve.cpp
题解：
用筛法将满足a×p=b(p为质数)的a，b建一条边(注意方向),然后跑二分图最大匹配，也可以直接建两条边，然后结果除以二
答案为：点数-最大匹配