2017ACM暑期多校联合训练 - Team 2 1003 HDU 6047 Maximum Sequence （线段树）

题目链接

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
4
8 11 8 5
3 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;

分析：

每次需要从已有的a数组（a数组保存的就是本来的值减去下标）里面找到一个最大值，然后将这个值加入到a数组后面，接着在重复上面的过程。每次在找到一个值之后，线段树都要更新。

#include <iostream>
#include <stdio.h>
#include <algorithm>
#define lchild left,mid,root<<1
#define rchild mid+1,right,root<<1|1
#define inf 0x3f3f3f3f
using namespace std;

const int maxn = 600000;
const int mod = 1e9+7;
int Max[maxn<<2];
int a[maxn];
int b[maxn];

void push_up(int root)
{
    Max[root] = max(Max[root<<1],Max[root<<1|1]);
}
///构建线段树
void build(int left,int right,int root)
{
    if(left == right)
    {
        Max[root] = a[left];
        return;
    }
    int mid = (left+right)>>1;
    build(lchild);
    build(rchild);
    push_up(root);
}
int query(int L,int R,int left,int right,int root)///[L,R]是需要查找的区间
{
    if(L<=left && right<=R)
        return Max[root];
    int mid = (left+right)>>1;
    int ans = -inf;
    if(L<=mid) ans = max(ans,query(L,R,lchild));
    if(R>mid) ans = max(ans,query(L,R,rchild));
    return ans;
}
void Insert(int pos,int left,int right,int root)
{
    if(left == right)
    {
        Max[root] = a[pos]-pos;
        return;
    }
    int mid = (left+right)>>1;
    if(pos<=mid) Insert(pos,lchild);
    else Insert(pos,rchild);
    push_up(root);
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i = 1; i <= n; i++)
        {
            scanf("%d",&a[i]);
            a[i] = a[i]-i;///a数组中直接保存为它本身减去下标的值，从而每次从里面取得最大值
        }
        for(int i = 1; i <= n; i++)
        {
            scanf("%d",&b[i]);
            a[i+n] = -inf;
        }
        sort(b,b+n);
        build(1,2*n,1);///因为在从后面取的时候要从整个数组里面取包括后面的部分
        int ans = 0;
        int range = n;
        for(int i = 1; i <= n; i++)
        {
            int num = query(b[i],range,1,2*n,1);
            range++;
            a[i+n] = num;
            Insert(n+i,1,2*n,1);
            ans = (ans+num)%mod;
        }
        printf("%d\n",ans);
    }
    return 0;
}