洛谷 P1231 教辅的组成 （三分图匹配，裂点）

题目

戳我

思路

三分图匹配，不过看上去裸匈牙利应该能过吧,我们考虑网络流，但是由于一本书只能匹配一次，那么我们考虑裂点来做流量限制，然后直接跑dinic就好了。

代码实现

#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
#define rep(i,f_start,f_end) for (int i=f_start;i<=f_end;++i)
#define per(i,n,a) for (int i=n;i>=a;i--)
#define MT(x,i) memset(x,i,sizeof(x) )
#define rev(i,start,end) for (int i=start;i<end;i++)
#define inf 0x3f3f3f3f
#define mp(x,y) make_pair(x,y)
#define lowbit(x) (x&-x)
#define MOD 1000000007
#define exp 1e-8
#define N 1000005 
#define fi first 
#define se second
#define pb push_back
typedef long long ll;
typedef pair<int ,int> PII;
typedef pair<int ,PII> PIII;
ll gcd (ll a,ll b) {return b?gcd (b,a%b):a; }
inline int read() {
    char ch=getchar(); int x=0, f=1;
    while(ch<'0'||ch>'9') {
        if(ch=='-') f=-1;
        ch=getchar();
    } while('0'<=ch&&ch<='9') {
        x=x*10+ch-'0';
        ch=getchar();
    } return x*f;
}
const int maxn=5e5+10;
int head[maxn],cnt=1;
struct edge {
	int v,flow,next;
}e[maxn];

inline void add (int u,int v,int flow) {
	e[++cnt]= (edge) {v,flow,head[u]};
	head[u]=cnt;
}

inline void add_edge (int u,int v,int flow) {
    add (u,v,flow);
	add (v,u,0);
}   

int dis[maxn],cur[maxn];
int s,t,n,m,c1,c2;
int bfs () {
   MT (dis,0);
   queue <int > q;
   dis[s]=1;
   q.push (s);
   while (q.size ()) {
      int x=q.front (); q.pop ();
	  for (int i=head[x];~i;i=e[i].next) {
		  if (dis[e[i].v]==0&&e[i].flow) {
			  dis[e[i].v]=dis[x]+1;
			  q.push (e[i].v);
		  }
	  }
   }
   return dis[t];
}

int dfs  (int now,int nowflow) {
	if (now==t) return nowflow;
	for (int &i=cur[now];i!=-1;i=e[i].next) {
        if (dis[e[i].v]==dis[now]+1&&e[i].flow) {
			int canflow=dfs (e[i].v,min (e[i].flow,nowflow));
			if (canflow) {
				e[i].flow-=canflow;
				e[i^1].flow+=canflow;
				return canflow;
			}
		}
	} 
	return 0;
}

int ans=0;
int x,y;
void Dinic () {
	while (bfs ()) {
		memcpy (cur,head,sizeof (cur));
		while (int val=dfs (s,inf)) ans+=val;
	}
}
int n1,n2,n3,m1,m2;
int main () {
	MT (head,-1);
    scanf ("%d%d%d",&n1,&n2,&n3);
	int num=2*n1+n2+n3;
	scanf  ("%d",&m1);
	s=0,t=num+1;
	rep (i,n3+1,n1+n3) add_edge (i,i+n1,1);
    
	rep (i,1,m1) {
		int x,y;
		scanf ("%d%d",&x,&y);
		add_edge (n3+x+n1,y+2*n1+n3,inf);
	}
	scanf ("%d",&m2);
	rep (i,1,m2) {
		scanf ("%d%d",&x,&y);
		add_edge (y,n3+x,inf);
	}
	rep (i,1,n3) add_edge (s,i,1);
	rep (i,1,n2) add_edge (2*n1+n3+i,t,1);
	Dinic ();
	cout<<ans<<endl;  
	return 0;
}