[PA2014]Muzeum

传送门

BZOJ

Solution

显然这是一个最大权闭合子图的问题,所以你把图建出来跑网络流就是(50pts).
接着你旋转坐标系然后把这个转换成为一个贪心替换网络流的问题,然后就是一个(set)的事了.

代码实现

/*
  mail: mleautomaton@foxmail.com
  author: MLEAutoMaton
  This Code is made by MLEAutoMaton
*/
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<iostream>
using namespace std;
#define ll long long
#define re register
#define file(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout)
inline int gi()
{
	int f=1,sum=0;char ch=getchar();
	while(ch>'9' || ch<'0'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0' && ch<='9'){sum=(sum<<3)+(sum<<1)+ch-'0';ch=getchar();}
	return f*sum;
}
const int N=5010,Inf=1e9+10;
struct thing
{
	int x,y,v;
}p[N],q[N];
int n,m,w,h,front[N],cnt,s,t,dep[N];
struct node
{
	int to,nxt,w;
}e[6000010];
void Add(int u,int v,int w){e[cnt]=(node){v,front[u],w};front[u]=cnt++;e[cnt]=(node){u,front[v],0};front[v]=cnt++;}
queue<int>Q;
bool bfs()
{
	Q.push(s);memset(dep,0,sizeof(dep));dep[s]=1;
	while(!Q.empty())
	{
		int u=Q.front();Q.pop();
		for(int i=front[u];i!=-1;i=e[i].nxt)
		{
			int v=e[i].to;
			if(!dep[v] && e[i].w)
			{
				dep[v]=dep[u]+1;
				Q.push(v);
			}
		}
	}
	return dep[t];
}
int dfs(int u,int flow)
{
	if(u==t || !flow)return flow;
	for(int i=front[u];i!=-1;i=e[i].nxt)
	{
		int v=e[i].to;
		if(dep[v]==dep[u]+1 && e[i].w)
		{
			int di=dfs(v,min(flow,e[i].w));
			if(di)
			{
				e[i].w-=di;e[i^1].w+=di;
				return di;
			}
			else dep[v]=0;
		}
	}
	return 0;
}
int Dinic()
{
	int flow=0;
	while(bfs())
		while(int d=dfs(s,Inf))
			flow+=d;
	return flow;
}
bool bein(int i,int j){return 1ll*abs(p[j].x-q[i].x)*h<=1ll*w*abs(p[j].y-q[i].y);}
namespace Accepted
{
	#define int ll
	const int N=500010;
	struct node
	{
		int x,y,v;
		bool operator<(const node b)const{return x<b.x;}
	}p[N],q[N];
	typedef pair<int,int> pii;
	set<pii>se;
	#define mp make_pair
	void main()
	{
		n=gi();m=gi();w=gi();h=gi();int ans=0;
		for(int i=1;i<=n;i++){int x=1ll*gi()*h,y=1ll*gi()*w,v=gi();p[i].x=x+y;p[i].y=x-y;p[i].v=v;ans+=v;}
		for(int i=1;i<=m;i++){int x=1ll*gi()*h,y=1ll*gi()*w,v=gi();q[i].x=x+y;q[i].y=x-y;q[i].v=v;}
		sort(p+1,p+n+1);	sort(q+1,q+m+1);
		int pos=1;
		for(int i=1;i<=m;i++)
		{
			while(pos<=n && p[pos].x<=q[i].x)se.insert(mp(p[pos].y,p[pos].v)),pos++;
			set<pii>::iterator it=se.lower_bound(mp(q[i].y,0));int flow=q[i].v;
			while(flow && it!=se.end())
			{
				pii now=*it;se.erase(it);
				int d=min(flow,now.second);
				now.second-=d;ans-=d;flow-=d;
				if(now.second)se.insert(now);
				else it=se.lower_bound(mp(q[i].y,0));
			}
		}
		printf("%lld
",ans);
		return;
	}
}
signed main()
{/*
	ll ans=0;
	n=gi();m=gi();memset(front,-1,sizeof(front));s=0;t=n+m+1;
	w=gi();h=gi();
	for(int i=1;i<=n;i++)p[i].x=gi(),p[i].y=gi(),p[i].v=gi(),Add(s,i,p[i].v),ans+=p[i].v;
	for(int i=1;i<=m;i++)q[i].x=gi(),q[i].y=gi(),q[i].v=gi(),Add(i+n,t,q[i].v);
	for(int i=1;i<=m;i++)
		for(int j=1;j<=n;j++)
			if(q[i].y>=p[j].y && bein(i,j))Add(j,i+n,Inf);
			printf("%lld
",ans-Dinic());*/
	Accepted::main();
	return 0;
}