CF837G Functions On The Segments

Functions On The Segments - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

题目大意

你有 $n$ 个函数，第 $i$ 个函数 $f_i$ 为：

$$f_i(x)=\{\begin{array}{ll}{y}_1,& x\le x_1\\ ax+b,& x_1\le x\le x_2\\ y_2,& x>x_2\end{array}$$

$m$ 次询问，每次询问给出 $l,r,x$，求 $\sum _{i=l}^r{f}_i(x)$。强制在线。

思路

主席树

转化一下式子就是：
$$f_i(x)=\{\begin{array}{ll}0x+y_1,& x\le x_1\\ ax+b,& x_1\le x\le x_2\\ 0x+y_2,& x>x_2\end{array}$$
所以答案就是：
$$\sum _{i=l}^r{a}_{i}x+\sum _{i=l}^r{b}_i$$
把 $x_1,x_1+1,x_2,x_2+1$ 离散化一下，然后开一棵主席树维护 $a$ 和 $b$ 的差分数组。

用差分是因为强制在线

code

#include <bits/stdc++.h>
#define fu(x , y , z) for(int x = y ; x <= z ; x ++)
#define LL long long
using namespace std;
const int N = 75005;
const LL maxx = INT_MAX;
int x[N] , xx[N] , a[N] , b[N] , n , cnt , rt[N * 2] , re[N * 8];
LL y[N] , yy[N];
struct Tr {int lp , rp , flg1 , flg2;LL v1 , v2;
} tr[N * 200];
void copy_tr (int p , int lst) {tr[p].v1 = tr[lst].v1;tr[p].v2 = tr[lst].v2;tr[p].lp = tr[lst].lp;tr[p].rp = tr[lst].rp;
}
void glp (int p) {if (tr[p].flg1 == 0) {int lst = tr[p].lp;tr[p].lp = ++cnt;copy_tr (cnt , lst);}tr[p].flg1 = 1;
}
void grp (int p) {if (tr[p].flg2 == 0) {int lst = tr[p].rp;tr[p].rp = ++cnt;copy_tr (cnt , lst);}tr[p].flg2 = 1;
}
void change (int p , int l , int r , int X , LL val , int flg) {if (flg == 1) tr[p].v1 += val;else tr[p].v2 += val;if (l == r)return;else {int mid = l + r >> 1;if (X <= mid) {glp(p);change (tr[p].lp , l , mid , X , val , flg);}else {grp (p);change (tr[p].rp , mid + 1 , r , X , val , flg);}}
}
struct node {LL a , b;
};
node query (int p , int l , int r , int L , int R) {if (L <= l && R >= r) return (node){tr[p].v1 , tr[p].v2};else {int mid = l + r >> 1;node ans1 = (node){0 , 0} , ans2 = (node){0 , 0};if (L <= mid && tr[p].lp)ans1 = query (tr[p].lp , l , mid , L , R);if (mid < R && tr[p].rp)ans2 = query (tr[p].rp , mid + 1 , r , L , R);return (node){ans1.a + ans2.a , ans1.b + ans2.b};}
}
int main () {scanf ("%d" , &n);int n1 = 0;re[++n1] = 0;re[++n1] = maxx;fu (i , 1 , n) {scanf ("%d%d%lld%d%d%lld" , &x[i] , &xx[i] , &y[i] , &a[i] , &b[i] , &yy[i]);re[++n1] = x[i];re[++n1] = xx[i];re[++n1] = xx[i] + 1;re[++n1] = x[i] + 1;}int T;sort (re + 1 , re + n1 + 1);int m = unique (re + 1 , re + n1 + 1) - re - 1;int aa , bb;fu (i , 1 , n) {rt[i] = ++cnt;copy_tr (rt[i] , rt[i - 1]); change (rt[i] , 1 , m , 1 , y[i] , 2);aa = upper_bound(re + 1 , re + m + 1 , x[i]) - re;change (rt[i] , 1 , m , aa , -y[i] , 2);aa = upper_bound(re + 1 , re + m + 1 , x[i]) - re , bb = upper_bound(re + 1 , re + m + 1 , xx[i]) - re;change (rt[i] , 1 , m , aa , a[i] , 1);change (rt[i] , 1 , m , aa , b[i] , 2);change (rt[i] , 1 , m , bb , -a[i] , 1);change (rt[i] , 1 , m , bb , -b[i] , 2);aa = upper_bound(re + 1 , re + m + 1 , xx[i]) - re;change (rt[i] , 1 , m , aa , yy[i] , 2);}node ans1 , ans2;LL ans3 , ans4 , lst = 0;int u , ll , rr;scanf ("%d" , &T);while (T --) {  scanf ("%d%d%d" , &ll , &rr , &u);        u = (u + lst) % 1000000000;aa = upper_bound(re + 1 , re + m + 1 , u) - re - 1;ans1 = query (rt[ll - 1] , 1 , m , 1 , aa);ans2 = query (rt[rr] , 1 , m , 1 , aa);ans3 = ans1.a * u + ans1.b;ans4 = ans2.a * u + ans2.b;lst = ans4 - ans3;printf ("%lld\n" , lst);}return 0;
}