HDU 6047 - Maximum Sequence | 2017 Multi-University Training Contest 2

/*
HDU 6047 - Maximum Sequence [ 单调队列 ]
题意：
	起初给出n个元素的数列 A[N], B[N]
	对于 A[]的第N+K个元素，从B[N]中找出一个元素B[i]，在 A[] 中找到一个数字A[p]满足 B[i] <= p <= N+K-1
	令 A[N+K] = A[p]-p，直到A[]的长度等于2N
	问 A[N+1] + A[N+2] + ... + A[N<<1] 最大是多少
分析：
	将A[]中元素全部减去其下标
	将B[]排序，可分析一定是从小往大选择最优
	每次从可以选择的A[]中选最大的一个，过程用单调队列来实现
编码时长：13分钟
*/
#include <bits/stdc++.h>
using namespace std;
#define LL long long
const LL MOD = 1e9+7;
const int N = 250004;
int n;
int a[N<<1], b[N];
struct Node {
    int x, id;
}q[N<<1];
int head, tail;
void solve()
{
    for (int i = 1; i <= n; i++) a[i] -= i;
    head = tail = 0;
    for (int i = 1; i <= n; i++)
    {
        while (head < tail && q[tail-1].x < a[i]) tail--;
        q[tail++] = Node{a[i], i};
    }
    for (int i = n+1; i <= n<<1; i++)
    {
        while (head < tail && q[head].id < b[i-n]) head++;
        a[i] = q[head].x-i;
        while (head < tail && q[tail-1].x < a[i]) tail--;
        q[tail++] = Node{a[i], i};
    }
}
int main()
{
    while (~scanf("%d", &n))
    {
        for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
        for (int i = 1; i <= n; i++) scanf("%d", &b[i]);
        sort(b+1, b+1+n);
        solve();
        LL ans = 0;
        for (int i = n+1; i <= n<<1; i++)
        {
            ans += a[i] + i;
            if (ans > MOD) ans -= MOD;
            if (ans < 0) ans += MOD;
        }
        ans = (ans%MOD + MOD) % MOD;
        printf("%lld\n", ans);
    }
}