What the hell is the bzoj question... The general idea of the title
: there is a sequence with an initial value of 0, n operations, each time the number in the interval of the sequence (ai, bi-1) and ci take max, after n times of asking The element sum
is discretized, and then a line segment tree is established, and each modification can be marked with a max mark on the interval.
#include<iostream>
#include<cstdio>
#include<map>
#include<algorithm>
using namespace std;
const int N=100005;
int n,g[N],tot,a[N],b[N],c[N],has,h[N];
map<int,int>mp;
struct qwe
{
int l,r,mx;
}t[N<<2];
int read()
{
int r=0,f=1;
char p=getchar();
while(p>'9'||p<'0')
{
if(p=='-')
f=-1;
p=getchar();
}
while(p>='0'&&p<='9')
{
r=r*10+p-48;
p=getchar();
}
return r*f;
}
void build(int ro,int l,int r)
{
t[ro].l=l,t[ro].r=r;
if(l==r)
return;
int mid=(l+r)>>1;
build(ro<<1,l,mid);
build(ro<<1|1,mid+1,r);
}
void pd(int ro)
{
if(t[ro].mx!=0)
{
t[ro<<1].mx=max(t[ro<<1].mx,t[ro].mx);
t[ro<<1|1].mx=max(t[ro<<1|1].mx,t[ro].mx);
}
}
void update(int ro,int l,int r,int v)
{
if(l>r)
return;
if(t[ro].l==l&&t[ro].r==r)
{
t[ro].mx=max(t[ro].mx,v);
return;
}
pd(ro);
int mid=(t[ro].l+t[ro].r)>>1;
if(r<=mid)
update(ro<<1,l,r,v);
else if(l>mid)
update(ro<<1|1,l,r,v);
else
update(ro<<1,l,mid,v),update(ro<<1|1,mid+1,r,v);
}
int ques(int ro,int p)
{
if(t[ro].l==t[ro].r)
return t[ro].mx;
pd(ro);
int mid=(t[ro].l+t[ro].r)>>1;
if(p<=mid)
return ques(ro<<1,p);
else
return ques(ro<<1|1,p);
}
int main()
{
n=read();
for(int i=1;i<=n;i++)
a[i]=read(),b[i]=read(),c[i]=read(),g[++tot]=a[i],g[++tot]=b[i];
sort(g+1,g+1+tot);
for(int i=1;i<=tot;i++)
if(i==1||g[i]!=g[i-1])
mp[g[i]]=++has,h[has]=g[i];
build(1,1,has);
for(int i=1;i<=n;i++)
update(1,mp[a[i]]+1,mp[b[i]],c[i]);
long long ans=0;
for(int i=2;i<=has;i++)
{
int nw=ques(1,i);
ans+=1ll*nw*(h[i]-h[i-1]);
}
printf("%lld\n",ans);
return 0;
}