UVA 11159 Factors 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 << endl
const 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;
}