HDU6047 Maximum Sequence（2017多校第2场）

Maximum Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 599    Accepted Submission(s): 306

Problem Description

Steph is extremely obsessed with “sequence problems” that are usually seen on magazines: Given the sequence 11, 23, 30, 35, what is the next number? Steph always finds them too easy for such a genius like himself until one day Klay comes up with a problem and ask him about it.

Given two integer sequences {ai} and {bi} with the same length n, you are to find the next n numbers of {ai}: an+1…a2n. Just like always, there are some restrictions on an+1…a2n: for each number ai, you must choose a number bk from {bi}, and it must satisfy ai≤max{aj-j│bk≤j<i}, and any bk can’t be chosen more than once. Apparently, there are a great many possibilities, so you are required to find max{∑2nn+1ai} modulo 109+7 .

Now Steph finds it too hard to solve the problem, please help him.

 

Input

The input contains no more than 20 test cases.
For each test case, the first line consists of one integer n. The next line consists of n integers representing {ai}. And the third line consists of n integers representing {bi}.
1≤n≤250000, n≤a_i≤1500000, 1≤b_i≤n.

 

Output

For each test case, print the answer on one line: max{∑2nn+1ai} modulo 109+7。

 

Sample Input

48 11 8 53 1 4 2

 

Sample Output

27
Hint
For the first sample:1. Choose 2 from {bi}, then a_2…a_4 are available for a_5, and you can let a_5=a_2-2=9; 2. Choose 1 from {bi}, then a_1…a_5 are available for a_6, and you can let a_6=a_2-2=9;
 

 

Source

2017 Multi-University Training Contest - Team 2

 

Recommend

liuyiding

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题目的意思是给出一个a序列和b序列，构造a序列后n项，构造方法是取b的一个数，选a序列从[b[i],j)的数减去索引的最大值，j是当前构造的位置。

思路：贪心b优先选最小的，线段树维护区间最大值，或者优先队列

线段树：

#include <iostream>  #include <cstdio>  #include <cstring>  #include <string>  #include <algorithm>  #include <map>  #include <cmath>#include <set>  #include <stack>  #include <queue>  #include <vector>  #include <bitset>  #include <functional>using namespace std;#define LL long long  const int INF = 0x3f3f3f3f;const LL mod = 1000000007;int n;int ma[2000009];int b[250009];void build(int k, int l, int r){    if (l == r)    {        if (l > n) return;        scanf("%d", &ma[k]); ma[k] -= l;        return;    }    int mid = (l + r) >> 1;    build(k << 1, l, mid);    build(k << 1 | 1, mid + 1, r);    ma[k] = max(ma[k << 1], ma[k << 1 | 1]);}int query(int k, int l, int r, int ll, int rr){    if (l >= ll&&r <= rr) return ma[k];    int ma = -1, mid = (l + r) >> 1;    if (mid >= ll) ma = max(ma, query(k << 1, l, mid, ll, rr));    if (rr > mid) ma = max(ma, query(k << 1 | 1, mid + 1, r, ll, rr));    return ma;}void update(int k, int l, int r, int p, int val){    if (l == r) { ma[k] = val; return; }    int mid = (l + r) >> 1;    if (p <= mid) update(k << 1, l, mid, p, val);    else update(k << 1 | 1, mid + 1, r, p, val);    ma[k] = max(ma[k << 1], ma[k << 1 | 1]);}int main(){    while (~scanf("%d", &n))    {        build(1, 1, 2 * n);        for (int i = 1; i <= n; i++) scanf("%d", &b[i]);        sort(b + 1, b + 1 + n);        LL ans = 0;        for (int i = n + 1; i <= 2 * n; i++)        {            int k = query(1, 1, 2 * n, b[i - n], i - 1);            update(1, 1, 2 * n, i, k - i);            ans += 1LL*k;            ans %= mod;        }        printf("%lld\n", ans);    }    return 0;}

优先队列：

#include <stdio.h>#include <stdlib.h>#include <string.h>#include <string>#include <math.h>#include <time.h>#include <algorithm>#include <complex>#include <vector>#include <stack>#include <queue>#include <unordered_map>#include <set>using namespace std;#define LL long long#define mem(a,b) memset(a,b,sizeof a);const int inf=0x3f3f3f3f;const LL llinf=0x3f3f3f3f3f3f3f3f;const double pi=acos(-1.0);const double eps=1e-8;const LL mod=1e9+7;const int maxn=2010;int a[250005],b[250005];struct node{    int id,val;    friend bool operator<(node a,node b)    {        return a.val<b.val;    }};int main(){    int n;    node f;    while(~scanf("%d",&n))    {        priority_queue<node>q;        for(int i=1; i<=n; i++)        {            scanf("%d",&a[i]),            f.id=i;            f.val=a[i]-i;                  q.push(f);        }        for(int i=1; i<=n; i++)            scanf("%d",&b[i]);        sort(b+1,b+n+1);        LL ans=0;        for(int i=1; i<=n; i++)        {            while(q.top().id<b[i])            {                q.pop();            }            f.id=n+i;            f.val=q.top().val-f.id;            ans+=q.top().val;            ans%=mod;            q.push(f);        }        printf("%lld\n",ans);    }    return 0;}