Va 12715 watching the kangaroo (Binary)

Question: N line segments (n <= 100000) (L <= r <= 1e9), M groups (M <= 100000) each time they ask about the maximum coverage of a point, for a line segment containing a vertex x, the coverage range is Min (X-L, R-x)

Idea: consider dividing a line segment into two parts. For the left part, sort the parts by the right endpoint, and then find the point that the right endpoint just meets as ID in the second part, the next step is to find the smallest vertex of L in the range of [ID, N], which can be obtained by obtaining the largest prefix. The maximum distance on the left is seg [ID]. L-prelmax [ID]. The maximum distance on the right is similar.

#include <iostream>#include <cstdio>#include <cstring>#include <vector>#include <string>#include <algorithm>#include <queue>#include <set>#include <map>using namespace std;typedef long long LL;const int maxn = 100000+10;#define REP(_,a,b) for(int _ = (a); _ <= (b); _++)struct seg{    int L,R;    seg(int L = 0,int R = 0):L(L),R(R){}};bool cmp1(seg a,seg b) {    if(a.R != b.R)  return a.R < b.R;    else return a.L < b.L;}bool cmp2(seg a,seg b) {    if(a.L != b.L)  return a.L < b.L;    else return a.R < b.R;}vector<seg>lft,rgt;int preLMax[maxn],preRMax[maxn];int n,m;int main(){    int ncase,T=1;    cin >> ncase;    while(ncase--) {        lft.clear();        rgt.clear();        scanf("%d%d",&n,&m);        for(int i = 0; i < n; i++) {            int L,R;            scanf("%d%d",&L,&R);            int mid = (L+R)>>1;            lft.push_back(seg(L,mid));            rgt.push_back(seg(mid,R));        }        sort(lft.begin(),lft.end(),cmp1);        preLMax[lft.size()] = 1e9;        for(int i = lft.size()-1; i >= 0; i--) {            preLMax[i] = min(preLMax[i+1],lft[i].L);        }        sort(rgt.begin(),rgt.end(),cmp2);        preRMax[0] = rgt[0].R;        for(int i = 1; i < rgt.size(); i++) {            preRMax[i] = max(preRMax[i-1],rgt[i].R);        }        printf("Case %d:\n",T++);        while(m--) {            int x,ans=0;            scanf("%d",&x);            int L = 0,R = lft.size()-1;            while(L <= R) {                int mid = (L+R) >>1;                if(lft[mid].R < x) {                    L = mid+1;                }else{                    R = mid-1;                }            }            ans = max(ans,x-preLMax[L]);            L = 0,R = rgt.size()-1;            while(L <= R) {                int mid = (L+R) >>1;                if(rgt[mid].L < x) {                    L = mid+1;                }else{                    R = mid-1;                }            }            ans = max(ans,preRMax[R]-x);            ans = max(ans,0);            printf("%d\n",ans);        }    }    return 0;}

Va 12715 watching the kangaroo (Binary)