二分图最大匹配

大家可以先看看这些大佬的博客（本人等过段时间自己写下自己的理解）;

https://www.cnblogs.com/shenben/p/5573788.html

http://blog.csdn.net/nice_punch/article/details/54957819

#include<bits/stdc++.h>
#define de(x) cout<<#x<<"="<<x<<endl;
#define dd(x) cout<<#x<<"="<<x<<" ";
#define rep(i,a,b) for(int i=a;i<(b);++i)
#define repd(i,a,b) for(int i=a;i>=(b);--i)
#define mt(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define inf 0x7f
#define pii pair<int,int>
#define pdd pair<double,double>
#define pdi pair<double,int>
#define mp(u,v) make_pair(u,v)
#define sz(a) a.size()
#define ull unsigned long long
#define ll long long
#define pb push_back
#define PI acos(-1.0)
const int mod = 1e9+7;
const int maxn = 1e3+5;
const double EPS = 1e-6;
using namespace std;
//左边是问题 右边是锦囊 
int gcd(int a,int b){
    return b==0?a:gcd(b,a%b);
}
bool vis[maxn];
int p[maxn];//p[i] 表示第i个锦囊的匹配的是哪个问题，负一就是都没有匹配 
vector<int> G[maxn];
bool fd(int u){
    int len = sz(G[u]);
    rep(i,0,len){
        int v = G[u][i];
        if(vis[v]) continue;
        vis[v] = true;
        if(p[v]==-1||fd(p[v])){ //如果当前这个锦囊没有问题匹配的话，或者匹配这个锦囊的问题可以找到新的锦囊的话 
            p[v] = u;//v这个锦囊匹配u这个问题 
            return true;
        }
    }
    return false;
}
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m)){
        //n种锦囊 m个问题 
        mt(p,-1);
        int x,y;
        rep(i,0,m){
            scanf("%d%d",&x,&y);
            G[i].pb(x);
            G[i].pb(y);
        }
        int ans = 0;
        rep(i,0,m){ 
            mt(vis,false);
            if(fd(i)) ans++;
            else break;//因为如果这题没有打出来 那他就不能继续往下答了 
        }
        printf("%d
",ans);
    }
    return 0;
}

//D. 唐纳德和他的数学老师
#include<bits/stdc++.h>
#define de(x) cout<<#x<<"="<<x<<endl;
#define dd(x) cout<<#x<<"="<<x<<" ";
#define rep(i,a,b) for(int i=a;i<(b);++i)
#define repd(i,a,b) for(int i=a;i>=(b);--i)
#define mt(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define inf 0x7f
#define pii pair<int,int>
#define pdd pair<double,double>
#define pdi pair<double,int>
#define mp(u,v) make_pair(u,v)
#define sz(a) a.size()
#define ull unsigned long long
#define ll long long
#define pb push_back
#define PI acos(-1.0)
const int mod = 1e9+7;
const int maxn = 1e6+5;
const int N = 3010;
const double EPS = 1e-6;
using namespace std;
int gcd(int a,int b){
    return b==0?a:gcd(b,a%b);
}
int a[maxn];
bool isprime[maxn];
int prime[maxn];
int tot;
vector<int> G[maxn];
bool vis[maxn];
int p[maxn];
void getprime(int mx){
    mt(isprime,true);mt(prime,0);
    tot = 0;
    isprime[0] = isprime[1] = false;
    rep(i,2,mx+1){
        if(isprime[i]) prime[++tot] = i;
        for(int j=1;j<=tot && i*prime[j]<=mx;j++){
            isprime[i*prime[j]] = false;
            if(!(i%prime[j])) break;
        }
    }
}
void getedge(int x){
    int u = x;
    for(int i=1;;i++){
        if(prime[i]*prime[i]>x) break;
        if(x%prime[i]==0) G[u].pb(prime[i]);
        while(x%prime[i]==0) x/=prime[i];
    }
    if(x>1) G[u].pb(x);
    return;
}
bool fd(int u){
    int len = sz(G[u]);
    rep(i,0,len){
        int v = G[u][i];
        if(vis[v]) continue;
        vis[v] = true;
        if(p[v]==-1||fd(p[v])){
            p[v] = u;
            return true;
        }
    }
    return false;
}
int main()
{
    getprime(maxn-1);
    int n;
    scanf("%d",&n);
    rep(i,0,n) {
        scanf("%d",&a[i]);
        getedge(a[i]);
    }
    int ans = 0;
    mt(p,-1);
    rep(i,0,n){
        mt(vis,false);
        if(fd(a[i])) ans++;
        else break;
    }
    printf("%d
",ans);
    return 0;
}