代码

#include <queue>#include <cstdio>#include <iostream>#define N 100001#define ls now << 1#define rs now << 1 | 1#define max(x, y) (p[x].a * 2 + p[x].b > p[y].a * 2 + p[y].b ? (x) : (y))int n, last, now, ans, M[N];std::priority_queue <int> q; struct node{	int a, b;}p[N];inline int read(){	int x = 0, f = 1;	char ch = getchar();	for(; !isdigit(ch); ch = getchar()) if(ch == ‘-‘) f = -1;	for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - ‘0‘;	return x * f;}inline void add(int x, int d){	for(; x; x -= x & -x) M[x] = max(M[x], d);}inline int query(int x){	int ret = 0;	for(; x <= n; x += x & -x) ret = max(ret, M[x]);	return ret;}int main(){	int i, j, x;	n = read();	for(i = 1; i <= n; i++) p[i].a = read();	for(i = 1; i <= n; i++) p[i].b = read();	for(i = 1; i <= n; i++) add(i, i);	last = 0;	for(i = 1; i <= n; i++)	{		now = query(last + 1);		if(q.empty() || (q.top() < (p[now].a - p[last].a) * 2 + p[now].b))		{			for(j = last + 1; j < now; j++) q.push(p[j].b);			ans += (p[now].a - p[last].a) * 2 + p[now].b;			last = now;		}		else		{			ans += q.top();			q.pop();		}		printf("%d\n", ans);	}	return 0;}