首页 > 代码库 > Light oj 1149 - Factors and Multiples 【二分图最大匹配】【好题】
Light oj 1149 - Factors and Multiples 【二分图最大匹配】【好题】
PDF (English) | Statistics | Forum |
Time Limit: 2 second(s) | Memory Limit: 32 MB |
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 mwill 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 |
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 |
#include <cstdio> #include <cstring> #include <cmath> #include <cstdlib> #include <queue> #include <stack> #include <vector> #include <map> #include <string> #include <algorithm> #define LL long long #define MAXN 100+10 #define MAXM 20000+10 #define INF 0x3f3f3f3f using namespace std; int a[110], b[110]; int Map[110][110]; bool used[110]; int pipei[110]; int N, M; int k = 1; int find(int x) { for(int i = 1; i <= N; i++) { if(!used[i] && Map[x][i]) { used[i] = true; if(pipei[i] == -1 || find(pipei[i])) { pipei[i] = x; return 1; } } } return 0; } void solve()//求匹配 { int ans = 0; memset(pipei, -1, sizeof(pipei)); for(int i = 1; i <= M; i++) { memset(used, false, sizeof(used)); ans += find(i); } printf("Case %d: ", k++); printf("%d\n", ans); } int main() { int t; scanf("%d", &t); while(t--) { memset(Map, 0, sizeof(Map)); scanf("%d", &N); for(int i = 1; i <= N; i++) scanf("%d", &a[i]); scanf("%d", &M); for(int i = 1; i <= M; i++) { scanf("%d", &b[i]); for(int j = 1; j <= N; j++) { if(b[i] % a[j] == 0)//整除关系建边 Map[i][j] = 1; } } solve(); } return 0; }
Light oj 1149 - Factors and Multiples 【二分图最大匹配】【好题】