出这题的人是怎么想出来的……
言归正传，这题是二分图套值域线段树。
首先经过 @Vfleaking的神奇建图后，把图拆成二分图，
不妨利用有向图最小割的性质建图（以前我一直以为最小割和边的方向无关，可这样的话很奇怪哦……）
理解悲剧……
我们可以利用边有向的性质解决黑白色块……
然后发现线段树很多……主席树闪亮登场
然后·就这麽A了？…………
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#include<cmath>
#include<cctype>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define RepD(i,n) for(int i=n;i>=0;i--)
#define MEM(a) memset(a,0,sizeof(a))
#define MEMI(a) memset(a,127,sizeof(a))
#define MEMi(a) memset(a,128,sizeof(a))
#define INF (2139062143)
#define MAXN (500000 +10)
#define MAXM (500000 +10)
int n,m,s,t;
int edge[MAXM],next[MAXM]={0},pre[MAXN]={0},weight[MAXM],size=1;
void addedge(int u,int v,int w)
{
edge[++size]=v;
weight[size]=w;
next[size]=pre[u];
pre[u]=size;
}
void addedge2(int u,int v,int w){/*cout<<u<<' '<<v<<' '<<w<<endl;*/addedge(u,v,w),addedge(v,u,0);}
int cnt[MAXN]={0},d[MAXN]={0};
long long totflow=0;
long long sap(int x,int flow)
{
if (x==t) return flow;
int nowflow=0;
Forp(x)
{
int &v=edge[p];
if (d[v]==d[x]-1&&weight[p])
{
long long fl=sap(v,min(weight[p],flow));
weight[p]-=fl,weight[p^1]+=fl,nowflow+=fl,flow-=fl;
if (!flow) return nowflow;
}
}
if (!(--cnt[d[x]++])) d[s]=t+1;
cnt[d[x]]++;
return nowflow;
}
struct node
{
node *ch[2];
node(){ch[0]=ch[1]=NULL;}
}q[MAXN],*root[MAXN]={NULL};
int tail=0;
void ins(node *&p,node *x,int l,int r,int c)
{
int m=(l+r) >>1;
p=&q[++tail];
if (x) *p=*x;
if (l==r) return;
if (c<=m) ins(p->ch[0],p->ch[0],l,m,c);
else ins(p->ch[1],p->ch[1],m+1,r,c);
}
void print(node *p)
{
//cout<<
}
node *ans[MAXN];
int ans_siz=0;
void qur(node *p,int l,int r,int L,int R)
{
if (!p||L>R) return;
int m=l+r >>1;
if (L<=l&&r<=R) {ans[++ans_siz]=p;return;}
if (l==r) return;
if (L<=m) qur(p->ch[0],l,m,L,R);
if (m<R) qur(p->ch[1],m+1,r,L,R);
}
int a2[MAXN],a[MAXN],b[MAXN],w[MAXN],l[MAXN],r[MAXN],p[MAXN];
int main()
{
freopen("bzoj3218.in","r",stdin);
// freopen(".out","w",stdout);
scanf("%d",&n);
For(i,n) scanf("%d%d%d%d%d%d",&a[i],&b[i],&w[i],&l[i],&r[i],&p[i]);
memcpy(a2,a,sizeof(a));
sort(a2+1,a2+n+1);
int size=unique(a2+1,a2+n+1)-(a2+1);
For(i,n)
{
a[i]=lower_bound(a2+1,a2+size+1,a[i])-(a2);
l[i]=lower_bound(a2+1,a2+size+1,l[i])-(a2);
r[i]=upper_bound(a2+1,a2+size+1,r[i])-(a2)-1;
}
For(i,n) ins(root[i],root[i-1],1,size,a[i]);
s=2*n+1; t=2*n+2;
For(i,n) addedge2(s,i,b[i]),addedge2(i,t,w[i]),addedge2(i,n+i,p[i]);
For(i,tail)
{
if (q[i].ch[0]) addedge2(t+i,t+((q[i].ch[0])-q),INF);
if (q[i].ch[1]) addedge2(t+i,t+((q[i].ch[1])-q),INF);
}
For(i,n)
{
ans_siz=0;
qur(root[i],1,size,a[i],a[i]);
qur(root[i-1],1,size,a[i],a[i]);
node *cur=ans[1],*last=NULL;
if (ans_siz>1) last=ans[2];
//cout<<t+((cur)-(q))<<'*';
addedge2(t+((cur)-(q)),i,INF);
if (ans_siz>1) addedge2(t+(cur-(q)),t+((last)-(q)),INF);
}
For(i,n)
{
ans_siz=0;
qur(root[i],1,size,l[i],r[i]);
For(j,ans_siz) addedge2(n+i,t+((ans[j])-(q)),INF);
}
//cout<<2*n+2+tail<<endl;
cnt[0]=2*n+2+tail;
while (d[s]<=t) {totflow+=sap(s,INF);}
//For(i,2*n+tail) cout<<d[i]<<' ';
long long ans=-totflow;
For(i,n) ans+=b[i]+w[i];
cout/*<<totflow<<' '*/<<ans<<endl;
return 0;
}