BZOJ3932 主席树

https://www.lydsy.com/JudgeOnline/problem.php?id=3932

题意：给出一些带有等级的线段，求一点上前K小个等级线段的等级之和

 

询问是对于每一个点询问前K小的和，比较容易想到的是对每一个点都建立一颗权值线段树，维护点的数量和点的等级和。

问题是空间太大，即使动态开点也远远不够，所以考虑用主席树来优化。

由于主席树和前缀和密切相关的特性，我们可以考虑用差分，每一个点T[i]代表这个点的权值线段树，对于一条线段，在S[i]出加上,E[i] + 1处减去，在建立主席树的同时就可以维护出所有点的权值线段树，直接查询即可。

 

注意：一看数据范围是要离散化的，但是如果去重之后，在权值线段树上取前K个就会遇到一个相同的点上有K + 1个这样的情况，这种情况可以考虑return sum / cnt * k,也可以选择不去重。

 

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x);  
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x);  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();}
while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;}
const double eps = 1e-9;
const int maxn = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7; 
int N,M,K;
LL Hash[maxn];
struct Task{
    LL S,E,P;
    Task(){}
    Task(LL S,LL E,LL P):S(S),E(E),P(P){}
}task[maxn];
vector<int>Q[2][maxn];
struct Tree{
    int lt,rt;
    LL cnt,sum;
}tree[maxn * 50];
int T[maxn],tot;
void newnode(int &t){
    t = ++tot;
    tree[t].cnt = tree[t].sum = 0;
}
void Build(int &t,int l,int r){
    newnode(t);
    if(l == r) return;
    int m = l + r >> 1;
    Build(tree[t].lt,l,m);
    Build(tree[t].rt,m + 1,r);
}
void update(int &t,int pre,int l,int r,LL p,LL flag){
    newnode(t);
    tree[t] = tree[pre];
    tree[t].sum += Hash[p] * flag; tree[t].cnt += flag;
    if(l == r) return;
    int m = l + r >> 1;
    if(p <= m) update(tree[t].lt,tree[pre].lt,l,m,p,flag);
    else update(tree[t].rt,tree[pre].rt,m + 1,r,p,flag);
}
LL query(int t,int l,int r,int k){
    if(l == r) return Hash[l];
    int m = l + r >> 1;
    int num = tree[tree[t].lt].cnt;
    if(num >= k) return query(tree[t].lt,l,m,k);
    else return query(tree[t].rt,m + 1,r,k - num) + tree[tree[t].lt].sum;
}
bool cmp(Task a,Task b){
    return a.P < b.P;
}
int main(){
    Sca2(M,N);
    for(int i = 1; i <= M ; i ++){
        LL S = read(),E = read(),P = read();
        task[i] = Task(S,E,P);
        Hash[i] = P;    
    }
    sort(Hash,Hash + 1 + M);
    sort(task + 1,task + 1 + M,cmp);
    for(int i = 1; i <= M ; i ++){
        LL S = task[i].S,E = task[i].E,P = task[i].P;
        Q[0][S].pb(i);
        Q[1][E + 1].pb(i);
    }
    Build(T[0],1,M);
    for(int i = 1; i <= N + 1; i ++){
        int pre = T[i] = T[i - 1];
        for(int j = 0 ; j < Q[0][i].size(); j ++){
            int v = Q[0][i][j];
            update(T[i],pre,1,M,v,1);
            pre = T[i];
        }
        for(int j = 0 ; j < Q[1][i].size(); j ++){
            int v = Q[1][i][j];
            update(T[i],pre,1,M,v,-1);
            pre = T[i];
        }
    }
    LL Pre = 1;
    for(int i = 1; i <= N ; i ++){
        LL X = read(),A = read(),B = read(),C = read();
        LL K = (A * Pre + B) % C + 1;
        if(K >= tree[T[X]].cnt) Pre = tree[T[X]].sum;
        else Pre = query(T[X],1,M,K);
        Prl(Pre);
    } 
    return 0;
}