2020杭电多校第四场 1007-Go Running dinic求二分图最大匹配

1007-Go Running

题意

给定平面上 \(n\) 个点，你可以选斜率为 \(1\) 与 \(−1\) 的直线去覆盖它，问最少要几条直线。

分析

对于平面上每个点有两条能覆盖它的直线，把这两条直线建点，这个点作为连接这两条直线的边，问题就转化为了二分图的最小点覆盖问题，二分图的最小点覆盖=二分图的最大匹配数，用\(dinic\)求二分图最大匹配可以在\(O(n\sqrt n)\)的时间复杂度解决。

Code

#include<algorithm>
#include<iostream>
#include<cstring>
#include<iomanip>
#include<sstream>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<cmath>
#include<stack>
#include<set>
#include<map>
#define rep(i,x,n) for(int i=x;i<=n;i++)
#define per(i,n,x) for(int i=n;i>=x;i--)
#define sz(a) int(a.size())
#define rson mid+1,r,p<<1|1
#define pii pair<int,int>
#define lson l,mid,p<<1
#define ll long long
#define pb push_back
#define mp make_pair
#define se second
#define fi first
using namespace std;
const double eps=1e-8;
const int mod=1e9+7;
const int N=2e5+10;
const int inf=1e9;
struct dinic{
	struct ppo{
		int v,c,next;
	}e[N<<2];
	int S,T,tot,h[N],d[N];
	void init(){
		S=0,T=1,tot=0;
		memset(h,-1,sizeof(h));
	}
	void ae(int u,int v,int c){
		e[tot]=ppo{v,c,h[u]};
		h[u]=tot++;
	}
	void add(int u,int v,int c){
		ae(u,v,c);
		ae(v,u,0);
	}
	bool bfs(){
		memset(d,-1,sizeof(d));
		queue<int>q;
		q.push(S);
		d[S]=0;
		while(!q.empty()){
			int u=q.front();q.pop();
			for(int i=h[u];~i;i=e[i].next){
				int x=e[i].v,w=e[i].c;
				if(d[x]==-1&&w){
					d[x]=d[u]+1;
					q.push(x);
				}
			}
		}
		return (d[T]!=-1);
	}
	int dfs(int u,int flow){
		if(u==T) return flow;
		int res=0;
		for(int i=h[u];~i;i=e[i].next){
			int x=e[i].v;
			if(e[i].c&&d[u]+1==d[x]){
				int tmp=dfs(x,min(flow,e[i].c));
				flow-=tmp;
				e[i].c-=tmp;
				e[i^1].c+=tmp;
				res+=tmp;
				if(flow==0) break;
			}
		}
		if(res==0) d[u]=-1;
		return res;
	}
	int solve(){
		int res=0;
		while(bfs()){
			res+=dfs(S,inf);
		}
		return res;
	}
}dick;
int T,n,a[N],b[N],c[N];
vector<int>g[N];
int gao(int a[]){
	rep(i,1,n) c[i]=a[i];
	sort(c+1,c+n+1);
	int tot=unique(c+1,c+n+1)-c-1;
	int mx=0;
	rep(i,1,n){
		a[i]=lower_bound(c+1,c+tot+1,a[i])-c+1;
		mx=max(a[i],mx);
	}
	return mx;
}
int main(){
	//ios::sync_with_stdio(false);
	//freopen("in","r",stdin);
	scanf("%d",&T);
	while(T--){
		dick.init();
		scanf("%d",&n);
		rep(i,1,n){
			int t,x;
			scanf("%d%d",&t,&x);
			a[i]=t+x;
			b[i]=t-x;
		}
		int n1=gao(a);
		int n2=gao(b)+n1;
		rep(i,1,n) b[i]+=n1;
		rep(i,1,n){
			g[0].pb(a[i]);
			g[b[i]].pb(1);
			g[a[i]].pb(b[i]);
		}
		rep(i,0,n2){
			sort(g[i].begin(), g[i].end());
			g[i].erase(unique(g[i].begin(), g[i].end()),g[i].end());
			for(int x:g[i]){
				dick.add(i,x,1);
			}
			g[i].clear();
		}
		printf("%d\n",dick.solve());
	}
	return 0;
}