HDU 4866 Shooting(持久化线段树)

view code//第二道持久化线段树，照着别人的代码慢慢敲,还是有点不理解
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long ll;
#define lson l,m,ls[rt]
#define rson m+1,r,rs[rt]
const int N = 200020;
const int M = N*40;
int T[M], ls[M], rs[M], sum[M], id[N];
ll val[M];
int n, m, X, P, tot;

struct seg
{
    int p, h, id;
    seg() {}
    seg(int p, int h, int id):p(p),h(h),id(id) {}
    bool operator < (const seg &o) const{
        return p<o.p || (p==o.p&&h>o.h);
    }
}e[N];
int ecnt;

struct node
{
    int h, id;
    node(int h=0, int id=0):h(h),id(id) {}
    bool operator < (const node &o) const
    {
        return h<o.h;
    }
}sto[N];

void build(int l, int r, int& rt)
{
    rt = ++tot;
    sum[rt] = val[rt]=0;
    if(l==r) return ;
    int m = (l+r)>>1;
    build(lson);
    build(rson);
}

void update(int p, int c, ll v, int last, int l, int r, int &rt)
{
    rt = ++tot;
    ls[rt] = ls[last], rs[rt] = rs[last];
     sum[rt]=sum[last]+c, val[rt] = val[last]+v;
    if(r==l) return ;
    int m = (l+r)>>1;
    if(p<=m) update(p, c, v, ls[last], lson);
    else update(p, c, v, rs[last], rson);
}

ll query(int k, int l, int r, int rt)
{
    if(k==0) return 0;
    if(l==r) return val[rt];
    int m = (l+r)>>1;
    ll ans =  0;
    if(sum[ls[rt]]>k) ans = query(k, lson);
    else ans = val[ls[rt]] + query(k-sum[ls[rt]], rson);
    return ans;
}

int find(int key) //找到最后一个小于key||(等于key&&高度大于0) 位置
{
    int l = 1, r = ecnt, ans=0;
    while(l<=r)
    {
        int m = (l+r)>>1;
        if(e[m].p<key || (e[m].p==key&&e[m].h>=0))
            ans = m, l = m+1;
        else r = m-1;
    }
//    int l=1,r=ecnt+1, ans;
//    while (l<r)
//    {
//        int mid=(l+r)>>1;
//        if (e[mid].p<key || (e[mid].p==key && e[mid].h>=0)){
//            ans=mid;
//            l=mid+1;
//        }
//        else{
//            r=mid;
//        }
//    }
    return ans;
}

void show()
{
    for(int i=1; i<ecnt; i++)
    {
        printf("e[%d] = {%d, %d, %d}\n", i, e[i].p, e[i].h, id[e[i].id]);
    }
}

int main()
{
//    freopen("in.txt", "r", stdin);
    while(scanf("%d%d%d%d", &n, &m, &X, &P)>0)
    {
        ecnt = 0, tot = 0;
        int l, r, w;
        for(int i=1; i<=n; i++)
        {
            scanf("%d%d%d", &l, &r, &w);
            e[++ecnt] = seg(l, w, i);
            e[++ecnt] = seg(r, -w, i);
            sto[i] = node(w, i);
        }
        sort(e+1, e+ecnt+1);
        sort(sto+1, sto+n+1);
        for(int i=1; i<=n; i++) id[sto[i].id]=i;

//        show();
        build(1, n, T[0]);
        for(int i=1; i<=ecnt; i++)
        {
            update(id[e[i].id], (e[i].h>=0?1:-1), e[i].h, T[i-1], 1, n, T[i]);
        }
        ll pre=1;
        int pos, a, b, c;
        while(m--)
        {
            scanf("%d%d%d%d", &pos, &a, &b, &c);
            ll k = (ll)(a*pre+b)%c;
            int p =find(pos);
//            printf("pos = %d, p = %d, k = %I64d\n", pos, p, k);
//            printf("sum[%d] = %I64d\n", p, val[T[p]]);
            ll ans = query(k, 1, n, T[p]);
            if(pre>P) ans *=2;
            pre = ans;
            printf("%I64d\n", ans);
        }
    }
    return 0;
}