uva 11159 Factors and Multiples
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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 setA. 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 (2 from set A and 1 from set B).
Input
The first line of input is an integer T(T<50) that determine the number of test cases. Each case consists of two lines. The first line starts with nfollowed by n integers. The second line starts with m followed by m integers. Both n and m will be in the range [1,100]. All the elements of the two sets will fit in 32 bit signed integer.
Output
For each case, output the case number followed by the answer.
Sample Input |
Output for Sample Input |
2 4 2 3 4 5 4 6 7 8 9 3 100 200 300 1 150 |
Case 1: 3 Case 2: 0 |
Problem Setter: Sohel Hafiz
Special Thanks: Jane Alam Jan
题目大意:
A 和 B 两个集合,从A里面删除一些数 K1 个,从 B里面删除一些数 K2 个,使得 B 中的任何一个元素 都不是 A中 的任意一个数的倍数。
解题思路:
二部图最大匹配,只要将最大匹配中的全部左边或右边点删掉即可。
解题代码:
#include <iostream> #include <cstring> #include <cstdio> #include <cmath> #include <set> #include <algorithm> using namespace std; const int maxn=110; int path[maxn][maxn],visited[maxn],link[maxn]; int n,m,a[maxn],b[maxn]; int can(int x){ for(int i=0;i<m;i++){ if(visited[i]==-1 && path[x][i]>0){ visited[i]=1; if( link[i]==-1 || can(link[i]) ){ link[i]=x; return 1; } } } return 0; } void solve(int casen){ int ans=0; for(int i=0;i<n;i++) for(int j=0;j<m;j++){ if(a[i]==0){ if(b[j]!=0) continue; else path[i][j]=1; } else if(b[j]%a[i]==0){ path[i][j]=1; } } for(int i=0;i<n;i++){ memset(visited,-1,sizeof(visited)); if(can(i)) ans++; } printf("Case %d: %d\n",casen,ans); } int main(){ //freopen("in.txt","r",stdin); int t; scanf("%d",&t); for(int k=1;k<=t;k++){ memset(path,0,sizeof(path)); memset(link,-1,sizeof(link)); scanf("%d",&n); for(int i=0;i<n;i++) scanf("%d",&a[i]); scanf("%d",&m); for(int i=0;i<m;i++) scanf("%d",&b[i]); solve(k); } return 0; }