HDU-4325 line segment tree + discretization

The time span of this question is too large. If we build the time directly, the memory is obviously insufficient, so we have to consider that there are only 10 ^ 5 flowers, so we need to adopt discretization, but if we only want to discretization the Time of the flowers, when we query a certain time point, the information in the online segment tree does not exist, so we cannot get the correct answer, therefore, we need to discretize the flowers and ask all the time, so that the information is complete.

The Code is as follows:

#include <cstdlib>#include <cstdio>#include <cstring>#include <algorithm>#include <map>using namespace std;int N, M, ti[300005], cnt, Q[100005];struct Flower{    int s, e;    }f[100005];map<int,int>mp;struct Node{    int l, r, lazy;    }e[1200005];void build(int p, int l, int r){    e[p].l = l, e[p].r = r;    e[p].lazy = 0;    if (l != r) {        int mid = (l + r) >> 1;        build(p<<1, l, mid);        build(p<<1|1, mid+1, r);    }}void push_down(int p){    if (e[p].lazy) {        e[p<<1].lazy += e[p].lazy;        e[p<<1|1].lazy += e[p].lazy;        e[p].lazy = 0;    }    }void modify(int p, int l, int r){    if (l == e[p].l && r == e[p].r) {        e[p].lazy += 1;        }    else {        push_down(p);        int mid = (e[p].l + e[p].r) >> 1;        if (r <= mid) {            modify(p<<1, l, r);        }        else if (l > mid) {            modify(p<<1|1, l, r);            }        else {            modify(p<<1, l, mid);            modify(p<<1|1, mid+1, r);            }    } }int query(int p, int pos){    if (e[p].l == e[p].r) {        return e[p].lazy;        }    else {        push_down(p);        int mid = (e[p].l + e[p].r) >> 1;        if (pos <= mid) {            return query(p<<1, pos);            }            else {            return query(p<<1|1, pos);            }    }}int main(){    int T, ca = 0;    scanf("%d", &T);    while (T--) {        cnt = -1;        mp.clear();        scanf("%d %d", &N, &M);        for (int i = 1; i <= N; ++i) {            scanf("%d %d", &f[i].s, &f[i].e);                ti[++cnt] = f[i].s, ti[++cnt] = f[i].e;        }        for (int i = 1; i <= M; ++i) {            scanf("%d", &Q[i]);            ti[++cnt] = Q[i];        }        sort(ti, ti+cnt+1);        cnt = unique(ti, ti+cnt+1) - ti;        for (int i = 0; i < cnt; ++i) {            mp[ti[i]] = i;        }        build(1, 0, cnt-1);        for (int i = 1; i <= N; ++i) {            modify(1, mp[f[i].s], mp[f[i].e]);        }        printf("Case #%d:\n", ++ca);        for (int i = 1; i <= M; ++i) {            printf("%d\n", query(1, mp[Q[i]]));        }    }    return 0;}