2017 ACM-ICPC 亚洲区(南宁赛区)网络赛 Frequent Subsets Problem
2017-09-25 19:39
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题目链接
Frequent Subsets Problem
1000ms
131072K
The frequent subset problem is defined as follows. Suppose UU={1,
2,\ldots…,N}
is the universe, and S_{1}S1, S_{2}S2,\ldots…,S_{M}SM are MM sets
over UU.
Given a positive constant \alphaα, 0<\alpha
\leq 10<α≤1,
a subset BB (B
\neq 0B≠0)
is α-frequent if it is contained in at least \alpha
MαM sets
of S_{1}S1, S_{2}S2,\ldots…,S_{M}SM,
i.e. \left
| \left \{ i:B\subseteq S_{i} \right \} \right | \geq \alpha M∣{i:B⊆Si}∣≥αM.
The frequent subset problem is to find all the subsets that are α-frequent. For example, let U=\{1,
2,3,4,5\}U={1,2,3,4,5}, M=3M=3, \alpha
=0.5α=0.5,
and S_{1}=\{1,
5\}S1={1,5}, S_{2}=\{1,2,5\}S2={1,2,5}, S_{3}=\{1,3,4\}S3={1,3,4}.
Then there are 33 α-frequent
subsets of UU,
which are \{1\}{1},\{5\}{5} and \{1,5\}{1,5}.
The first line contains two numbers NN and \alphaα,
where NN is
a positive integers, and \alphaα is
a floating-point number between 0 and 1. Each of the subsequent lines contains a set which consists of a sequence of positive integers separated by blanks, i.e., line i
+ 1i+1 contains S_{i}Si, 1
\le i \le M1≤i≤M .
Your program should be able to handle NN up
to 2020and MM up
to 5050.
The number of \alphaα-frequent
subsets.
2017
ACM-ICPC 亚洲区(南宁赛区)网络赛
题意:现在给你一个n,再给你不多于50组的数据(及大集合)(每一行算一组数据)(每组数据的数字个数未知,且无重复,且数字不大于n)
现在让你求出,有多少个不同的 子集出现在这些集合中的概率大于a。
思路:数据不大,n最大20,最多50组数据;
第几组数据 1 2 3 4 5 ... (最多50组)
1 : 1 0 0 0 0
2 : 1 0 1 1 1
3 : 0 1 0 1 1
4 : 1 0 1 0 1
5 : 0 0 0 0 1
6 : 0 1 0 0 0
7 : 0 1 0 1 0
8 : 1 0 1 0 1
9 : 0 0 0 1 0
10 : 0 0 1 0 0
11 : 0 1 0 0 0
12 : 0 0 0 1 0
13 : 1 0 0 0 0
14 : 1 0 0 0 0
15 : 0 0 0 1 0
n : ..............
压缩成n个LL数字(横着看的二进制)
然后枚举所有的可能的子集合,把这个子集合所有对应的LL数字相与(&)后得到的数字转换成的二进制中1的个数就是包含这个子集合的大集合的个数
例1:存在子集{2,4,8}的大集合有多少个?
(1 0 1 1 1 )& (1 0 1 0 1)&(1 0 1 0 1)=(1 0 1 0 1)代表1,3,5组大集合中含有子集{2,4,8};
例2:存在子集{3,7}的大集合有多少个?
(0 1 0 1 1 )& (0 1 0 1 0)=(0 1 0 1 0)代表2,4组大集合中含有子集{3,7};
代码:
另一种代码:
hash思想吧,把每组数据hash成一个数字。思想类似
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<cstdio>
#define ll long long
#define lz 2*u,l,mid
#define rz 2*u+1,mid+1,r
#define mset(a,x) memset(a,x,sizeof(a))
using namespace std;
const double PI=acos(-1);
const int inf=0x3f3f3f3f;
const double esp=1e-12;
const int maxn=400005;
const int mod=1e9+7;
int dir[4][2]={0,1,1,0,0,-1,-1,0};
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll inv(ll b){if(b==1)return 1; return (mod-mod/b)*inv(mod%b)%mod;}
ll fpow(ll n,ll k){ll r=1;for(;k;k>>=1){if(k&1)r=r*n%mod;n=n*n%mod;}return r;}
int a[101];
int main()
{
int n,x,i;
double k;
cin>>n>>k;
n=(1<<n);
mset(a,0);
int top=1;
while(scanf("%d",&x)!=EOF)
{
a[top]+=(1<<(x-1));
if(getchar()=='\n')
top++;
}
int ans=0;
for(i=1;i<n;i++)
{
int c=0;
for(int j=1;j<=top;j++)
{
if((a[j]&i)==i)
c++;
}
if(1.0*c/top>=k-esp)
ans++;
}
cout<<ans<<endl;
return 0;
}
Frequent Subsets Problem
1000ms
131072K
The frequent subset problem is defined as follows. Suppose UU={1,
2,\ldots…,N}
is the universe, and S_{1}S1, S_{2}S2,\ldots…,S_{M}SM are MM sets
over UU.
Given a positive constant \alphaα, 0<\alpha
\leq 10<α≤1,
a subset BB (B
\neq 0B≠0)
is α-frequent if it is contained in at least \alpha
MαM sets
of S_{1}S1, S_{2}S2,\ldots…,S_{M}SM,
i.e. \left
| \left \{ i:B\subseteq S_{i} \right \} \right | \geq \alpha M∣{i:B⊆Si}∣≥αM.
The frequent subset problem is to find all the subsets that are α-frequent. For example, let U=\{1,
2,3,4,5\}U={1,2,3,4,5}, M=3M=3, \alpha
=0.5α=0.5,
and S_{1}=\{1,
5\}S1={1,5}, S_{2}=\{1,2,5\}S2={1,2,5}, S_{3}=\{1,3,4\}S3={1,3,4}.
Then there are 33 α-frequent
subsets of UU,
which are \{1\}{1},\{5\}{5} and \{1,5\}{1,5}.
Input Format
The first line contains two numbers NN and \alphaα,where NN is
a positive integers, and \alphaα is
a floating-point number between 0 and 1. Each of the subsequent lines contains a set which consists of a sequence of positive integers separated by blanks, i.e., line i
+ 1i+1 contains S_{i}Si, 1
\le i \le M1≤i≤M .
Your program should be able to handle NN up
to 2020and MM up
to 5050.
Output Format
The number of \alphaα-frequentsubsets.
样例输入
15 0.4 1 8 14 4 13 2 3 7 11 6 10 8 4 2 9 3 12 7 15 2 8 3 2 4 5
样例输出
11
题目来源
2017ACM-ICPC 亚洲区(南宁赛区)网络赛
题意:现在给你一个n,再给你不多于50组的数据(及大集合)(每一行算一组数据)(每组数据的数字个数未知,且无重复,且数字不大于n)
现在让你求出,有多少个不同的 子集出现在这些集合中的概率大于a。
思路:数据不大,n最大20,最多50组数据;
第几组数据 1 2 3 4 5 ... (最多50组)
1 : 1 0 0 0 0
2 : 1 0 1 1 1
3 : 0 1 0 1 1
4 : 1 0 1 0 1
5 : 0 0 0 0 1
6 : 0 1 0 0 0
7 : 0 1 0 1 0
8 : 1 0 1 0 1
9 : 0 0 0 1 0
10 : 0 0 1 0 0
11 : 0 1 0 0 0
12 : 0 0 0 1 0
13 : 1 0 0 0 0
14 : 1 0 0 0 0
15 : 0 0 0 1 0
n : ..............
压缩成n个LL数字(横着看的二进制)
然后枚举所有的可能的子集合,把这个子集合所有对应的LL数字相与(&)后得到的数字转换成的二进制中1的个数就是包含这个子集合的大集合的个数
例1:存在子集{2,4,8}的大集合有多少个?
(1 0 1 1 1 )& (1 0 1 0 1)&(1 0 1 0 1)=(1 0 1 0 1)代表1,3,5组大集合中含有子集{2,4,8};
例2:存在子集{3,7}的大集合有多少个?
(0 1 0 1 1 )& (0 1 0 1 0)=(0 1 0 1 0)代表2,4组大集合中含有子集{3,7};
代码:
#include<stdio.h> #include<math.h> #include<string.h> #include<algorithm> #define LL long long using namespace std; int a[55][25]; LL d[25]; char s[100]; int ans=0; double ci; int LL1(LL x) { int sum=0; while(x) { if(x&1) sum++; x>>=1; } return sum; } void dfs(int x,int n,LL w) { if(LL1(w)<ci) return ; if(x>n) { ans++; return ; } dfs(x+1,n,w&d[x]); dfs(x+1,n,w); return ; } int main() { memset(a,0,sizeof(a)); memset(d,0,sizeof(d)); int n,k=0; double m; scanf("%d%lf",&n,&m); getchar(); while(gets(s)) { int la=strlen(s); int sum=0; for(int i=0; i<la; i++) { if(s[i]==' ') { a[k][sum]=1; sum=0; continue; } sum=sum*10+s[i]-'0'; } a[k][sum]=1; k++; } ci=k*m-0.0000001; for(int j=1; j<=n; j++) for(int i=0; i<k; i++) d[j]=(d[j]<<1)+a[i][j]; dfs(1,n,(1LL<<51)-1);//(1LL<<51)-1代表二进制50个1(初始化) printf("%d\n",ans-1); }
另一种代码:
hash思想吧,把每组数据hash成一个数字。思想类似
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<cstdio>
#define ll long long
#define lz 2*u,l,mid
#define rz 2*u+1,mid+1,r
#define mset(a,x) memset(a,x,sizeof(a))
using namespace std;
const double PI=acos(-1);
const int inf=0x3f3f3f3f;
const double esp=1e-12;
const int maxn=400005;
const int mod=1e9+7;
int dir[4][2]={0,1,1,0,0,-1,-1,0};
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll inv(ll b){if(b==1)return 1; return (mod-mod/b)*inv(mod%b)%mod;}
ll fpow(ll n,ll k){ll r=1;for(;k;k>>=1){if(k&1)r=r*n%mod;n=n*n%mod;}return r;}
int a[101];
int main()
{
int n,x,i;
double k;
cin>>n>>k;
n=(1<<n);
mset(a,0);
int top=1;
while(scanf("%d",&x)!=EOF)
{
a[top]+=(1<<(x-1));
if(getchar()=='\n')
top++;
}
int ans=0;
for(i=1;i<n;i++)
{
int c=0;
for(int j=1;j<=top;j++)
{
if((a[j]&i)==i)
c++;
}
if(1.0*c/top>=k-esp)
ans++;
}
cout<<ans<<endl;
return 0;
}
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