LightOJ--1149--Factors and Multiples(二分图好题)
2016-02-02 15:35
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Factors and Multiples
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Status
Description
You will be given two sets of integers. Let's call them set A and set
B. Set A contains n elements and set
B contains m elements. You have to remove
k1 elements from set A and k2 elements from set
B so that of the remaining values no integer in set B is a multiple of any integer in set
A. k1 should be in the range
[0, n] and k2 in the range [0, m].
You have to find the value of (k1 + k2) such that
(k1 + k2) is as low as possible. P is a multiple of
Q if there is some integer K such that
P = K * Q.
![](http://7xjob4.com1.z0.glb.clouddn.com/16029eeb2d7c5ee9486df4d79d590523)
Suppose set A is {2, 3, 4, 5} and set
B is {6, 7, 8, 9}. By removing 2 and
3 from A and 8 from B, we get the sets
{4, 5} and {6, 7, 9}. Here none of the integers
6, 7 or 9 is a multiple of 4 or
5.
So for this case the answer is 3 (two from set
A and one from set B).
Input
Input starts with an integer T (≤ 50), denoting the number of test cases.
The first line of each case starts with an integer n followed by
n positive integers. The second line starts with m followed by
m positive integers. Both n and m will be in the range
[1, 100]. Each element of the two sets will fit in a 32 bit signed integer.
Output
For each case of input, print the case number and the result.
Sample Input
2
4 2 3 4 5
4 6 7 8 9
3 100 200 300
1 150
Sample Output
Case 1: 3
Case 2: 0
Source
Problem Setter: Sohel Hafiz
Special Thanks: Jane Alam Jan
给了两个集合A,B,分别有n,m个数,从A取k1个数,B取k2个数,使得b[ j ]%a[ i ]==0的情况不存在
刚开始以为可以暴力的,但是后来发现暴力真的是挺麻烦,把图画出来之后会发现,其实就是最小点覆盖,二分图性质:最小点覆盖=最大匹配,匈牙利算法跑一次
Time Limit: 2000MS | Memory Limit: 32768KB | 64bit IO Format: %lld & %llu |
Status
Description
You will be given two sets of integers. Let's call them set A and set
B. Set A contains n elements and set
B contains m elements. You have to remove
k1 elements from set A and k2 elements from set
B so that of the remaining values no integer in set B is a multiple of any integer in set
A. k1 should be in the range
[0, n] and k2 in the range [0, m].
You have to find the value of (k1 + k2) such that
(k1 + k2) is as low as possible. P is a multiple of
Q if there is some integer K such that
P = K * Q.
Suppose set A is {2, 3, 4, 5} and set
B is {6, 7, 8, 9}. By removing 2 and
3 from A and 8 from B, we get the sets
{4, 5} and {6, 7, 9}. Here none of the integers
6, 7 or 9 is a multiple of 4 or
5.
So for this case the answer is 3 (two from set
A and one from set B).
Input
Input starts with an integer T (≤ 50), denoting the number of test cases.
The first line of each case starts with an integer n followed by
n positive integers. The second line starts with m followed by
m positive integers. Both n and m will be in the range
[1, 100]. Each element of the two sets will fit in a 32 bit signed integer.
Output
For each case of input, print the case number and the result.
Sample Input
2
4 2 3 4 5
4 6 7 8 9
3 100 200 300
1 150
Sample Output
Case 1: 3
Case 2: 0
Source
Problem Setter: Sohel Hafiz
Special Thanks: Jane Alam Jan
给了两个集合A,B,分别有n,m个数,从A取k1个数,B取k2个数,使得b[ j ]%a[ i ]==0的情况不存在
刚开始以为可以暴力的,但是后来发现暴力真的是挺麻烦,把图画出来之后会发现,其实就是最小点覆盖,二分图性质:最小点覆盖=最大匹配,匈牙利算法跑一次
#include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<algorithm> using namespace std; vector<int>map[200]; int used[200],pipei[200],a[200],b[200]; int n,m; int find(int x) { for(int i=0;i<map[x].size();i++) { int y=map[x][i]; if(!used[y]) { used[y]=1; if(pipei[y]==-1||find(pipei[y])) { pipei[y]=x; return 1; } } } return 0; } int main() { int t,k=1; scanf("%d",&t); while(t--) { memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); memset(pipei,-1,sizeof(pipei)); scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%d",&a[i]); map[i].clear(); } scanf("%d",&m); for(int i=0;i<m;i++) scanf("%d",&b[i]); for(int i=0;i<n;i++) { for(int j=0;j<m;j++) { if(b[j]%a[i]==0) { map[i].push_back(j); } } } int sum=0; for(int i=0;i<n;i++) { memset(used,0,sizeof(used)); sum+=find(i); } printf("Case %d: %d\n",k++,sum); } return 0; }
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