【luogu2756】 飞行员配对方案问题 [二分图匹配 匈牙利算法]

luogu2756

匈牙利 然后输出match就好了
我会说是因为我的最大流写这题写挂了我才来写匈牙利的吗

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
#define Min(x,y) ((x)<(y)?(x):(y))
const int N=200+50,M=30000+5,inf=0x3f3f3f3f,P=19650827;
int n,m,ans=0,match[N];
bool link[N][N],vis[N];
template <class t>void rd(t &x){
    x=0;int w=0;char ch=0;
    while(!isdigit(ch)) w|=ch=='-',ch=getchar();
    while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
    x=w?-x:x;
}

bool dfs(int x){
	for(int i=m+1;i<=n;++i)
	if(link[x][i]&&!vis[i]){
		vis[i]=1;
		if(!match[i]||dfs(match[i])){match[i]=x;return 1;}
	}
	return 0;
}

int main(){
	freopen("in2.txt","r",stdin);
	//freopen("xor.out","w",stdout);
	rd(m),rd(n);
	int x,y;
	while(scanf("%d%d",&x,&y)&&(x+y+2)) link[x][y]=1;
	for(int i=1;i<=m;++i){
		memset(vis,0,sizeof(vis));
		if(dfs(i)) ++ans;
	}
	printf("%d\n",ans);
	for(int i=m+1;i<=n;++i)
	if(match[i]) printf("%d %d\n",match[i],i);
	return 0;
}

63昏

#include
#include
#include
#include
#include
#include
using namespace std;
#define Min(x,y) ((x)<(y)?(x):(y))
const int N=200+50,M=30000+5,inf=0x3f3f3f3f,P=19650827;
int n,m,tt,s,t,ans,pre[N],ans1[N],ans2[N];
template void rd(t &x){
    x=0;int w=0;char ch=0;
    while(!isdigit(ch)) w|=ch=='-',ch=getchar();
    while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
    x=w?-x:x;
}
int head[N],tot=1;
struct edge{int v,flo,nxt;}e[M<<1];
void add(int u,int v,int flo){
	e[++tot]=(edge){v,flo,head[u]},head[u]=tot;
	e[++tot]=(edge){u,0,head[v]},head[v]=tot;
}
queueq;bool vis[N],used[N];
bool bfs(){
	while(!q.empty()) q.pop();
	memset(vis,0,sizeof(vis));
	q.push(s),vis[s]=1;
	while(!q.empty()){
		int u=q.front();q.pop();
		for(int i=head[u],v;i;i=e[i].nxt)
		if(e[i].flo&&!vis[v=e[i].v]){
			q.push(v),pre[v]=i,vis[v]=1;
			if(v==t) return 1;
		}
	}
	return 0;
}
void upd(){
	int x=t;
	while(x!=s){
		int i=pre[x];
		--e[i].flo,++e[i^1].flo;
		if(!used[x]){
		if(x<=m&&x) ans1[++ans1[0]]=x,used[x]=1;
		if(x>m&&x<=n) ans2[++ans2[0]]=x,used[x]=1;
		}
		x=e[i^1].v;
	}
	++ans;
}
int main(){
//	freopen("in2.txt","r",stdin);
	//freopen("xor.out","w",stdout);
	rd(m),rd(n);s=n+1,t=s+1;
	int x,y;ans=ans1[0]=ans2[0]=0;
	while(scanf("%d%d",&x,&y)!=EOF&&(x+y+2))  add(x,y,1);
	for(int i=1;i<=m;++i) add(s,i,1);
	for(int i=m+1;i<=n;++i) add(i,t,1);
	while(bfs())
	upd();
	printf("%d\n",ans);
	for(int i=1;i<=ans;++i) printf("%d %d\n",ans1[i],ans2[i]);
	return 0;
}