传送门
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;
}