[NOI2014] section

background

      In the beginning I look ignorant force, then next syk gods say "sorted by length", suddenly realized 

The meaning of problems

solution

    + Sorted by length discretization

      Then again sweep two points, each right end of a right, put the point corresponding to section +1 point for every right shift left, put the point corresponding to the interval -1 If the maximum interval> = m is feasible

code

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#define N 3000000
#define ls (x<<1)
#define rs (x<<1|1)
#define mid ((l+r)>>1)
using namespace std;
const int inf=2100000000;
int res[N<<3],lazy[N<<3];
int L[N],R[N],id[N],plk[N];
int n,m,tot=0;
void update(int x){res[x]=max(res[ls],res[rs]);}
void push_down(int x){
    if(lazy[x])lazy[ls]+=lazy[x],lazy[rs]+=lazy[x],res[ls]+=lazy[x],res[rs]+=lazy[x],lazy[x]=0;
}
void modify(int x,int l,int r,int nl,int nr,int k){
    push_down(x);
    if(nl<=l&&r<=nr){lazy[x]+=k;res[x]+=k;return;}
    if(mid>=nl)modify(ls,l,mid,nl,nr,k);
    if(mid<nr)modify(rs,mid+1,r,nl,nr,k);
    update(x);
}
bool cmp(int a,int b){return (R[a]-L[a])>(R[b]-L[b]);}
int main(){
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++){scanf("%d%d",L+i,R+i);plk[++tot]=L[i],plk[++tot]=R[i];}
for(int i=1;i<=n;i++)id[i]=i;
sort(id+1,id+n+1,cmp);
sort(plk+1,plk+tot+1);tot=unique(plk+1,plk+tot+1)-plk-1;
for(int i=1;i<=n;i++)L[i]=lower_bound(plk+1,plk+tot+1,L[i])-plk,R[i]=lower_bound(plk+1,plk+tot+1,R[i])-plk;//离散化
int pos1,pos2=0,ans=inf;
for(pos1=1;pos1<=n;pos1++){
    while(res[1]<m&&pos2<n){pos2++;modify(1,1,tot,L[id[pos2]],R[id[pos2]],1);}if(res[1]<m)break;
    years = min (years (plk [R [id [pos1]]] - l k [L [id [pos1]]]) - (l k [R [id [pos2]]] - l k [L [id [pos2] ]])); 
    modify ( 1 , 1 , tot, L [id [pos1]], R [id [pos1]] - 1 ); 
} // two point 
printf ( " % d \ n " , (== inf years) - 1 : years); //压行压行awa 
}