bzoj 2369 interval

http://www.elijahqi.win/archives/3213
Description

For an interval set
{A1, A2...Ak} (K>1, Ai is not equal to Aj (i is not equal to J), define its weight

S=|A1∪A2∪……AK|*|A1∩A2……∩Ak|
That is, the length of their intersection interval multiplied by the length of their union interval.
Obviously, if these intervals do not intersect, the weight is 0.
Your Task
gives you several unequal intervals, and selects several intervals to maximize their weights.
Input

The first line n represents the number of intervals, the
next n lines, two integers in each line lr describe an interval [l,r]
Output

Output maximum weights in one row
Sample Input

4
1 6
4 8
2 7
3 5
Sample Output

24
HINT

Example explanation

Choosing [1,6] and [2,7] is optimal.

data convention

100%：1< N<=10^6,1<=L< R<=10^6

Source

Suddenly I feel that the complexity is not very correct

Same as bzoj2687

https://blog.csdn.net/elijahqi/article/details/80077458

#include<cstdio>
#include<cctype> 
#include<algorithm>
#define ll long long 
using namespace std;
inline char gc(){
    static char now[1<<16],*S,*T;
    if (T==S){T=(S=now)+fread(now,1,1<<16,stdin);if (T==S) return EOF;}
    return *S++; 
}
inline int read(){
    int x=0,f=1;char ch=gc();
    while(!isdigit(ch)) {if (ch=='-') f=-1;ch=gc();}
    while(isdigit(ch)) x=x*10+ch-'0',ch=gc();
    return x*f;
}
const int N=1e6+10;
struct node{
    int l,r;
}line[N],t[N];
ll dp[N],ans;int n;
inline bool cmp(const node &a,const node &b){return a.l==b.l?a.r>b.r:a.l<b.l;};
inline void gao(int l,int r,int L,int R){
    if(l>r) return;
    if (L==R){
        for (int i=l;i<=r;++i) dp[i]=(ll)(line[i].r-line[L].l)*(line[L].r-line[i].l);return;
    }
    int mid=l+r>>1;int p=0;
    for (int i=L;i<=min(R,mid-1);++i){
        ll tmp=(ll)(line[mid].r-line[i].l)*(line[i].r-line[mid].l);
        if (tmp>dp[mid]) dp[mid]=tmp,p=i;
    }if (!p) gao(l,mid-1,L,mid-1),gao(mid+1,r,L,R);
    else gao(l,mid-1,L,p),gao(mid+1,r,p,R);
}
int main(){
    freopen("bzoj2369.in","r",stdin);
    n=read();int l=0,r=0,cnt=0;
    for (int i=1;i<=n;++i) t[i].l=read(),t[i].r=read();
    sort(t+1,t+n+1,cmp);
    for (int i=1;i<=n;++i){
        if (t[i].r>r) line[++cnt]=t[i],r=t[i].r,l=t[i].l;
        else ans=max(ans,(ll)(t[i].r-t[i].l)*(r-l));
    }
    gao(2,cnt,1,cnt);
    for (int i=1;i<=cnt;++i) ans=max(ans,dp[i]);
    printf("%lld\n",ans);
    return 0;
}