UVA 11159Factors and Multiples （最大独立集）

题意:

两个集合,如果B集合元素是A的倍数的话,就表明这俩有关系,问从这俩集合删除最少的元素使得没有关系

分析:

赤果果的二分图最大独立集=n−最大匹配,那么要删除的就是最大匹配数了
注意别模0了,这里Q=k∗P,由于k为整数也就是说0是任何数的倍数−−,因为0=0∗p

代码:

////  Created by TaoSama on 2015-11-07//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n, m, a[105], b[105];int g[105][105], match[105];bool vis[105];bool dfs(int u) {    for(int v = 1; v <= m; ++v) {        if(!g[u][v] || vis[v]) continue;        vis[v] = true;        if(!match[v] || dfs(match[v])) {            match[v] = u;            return true;        }    }    return false;}int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int t; scanf("%d", &t);    int kase = 0;    while(t--) {        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);        memset(g, 0, sizeof g);        for(int i = 1; i <= n; ++i)            for(int j = 1; j <= m; ++j)                if(a[i] && b[j] % a[i] == 0) g[i][j] = 1;                else if(!b[j]) g[i][j] = 1;        memset(match, 0, sizeof match);        int ans = 0;        for(int i = 1; i <= n; ++i) {            memset(vis, 0, sizeof vis);            ans += dfs(i);        }        printf("Case %d: %d\n", ++kase, ans);    }    return 0;}