HDU 6047 Maximum Sequence (贪心,线段树)

Maximum Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

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;

按照b的大小对b数组排序,贪心的选取然后更新.因为n的值连续越来越大,差值会越来越小,所以越大的数放在越前面越好.

#include <cmath>
#include <cstring>
#include <cstdio>
#include <vector>
#include <string>
#include <algorithm>
#include <string>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <ctime>
#include <cstdlib>
#include <iostream>
using namespace std;
const int INF=1e9+7;
const int MAXN=250010;
const int MODE=1e9+7;
int a[MAXN],b[MAXN];
int s[MAXN];
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
int MAX[MAXN<<2];
void pushup(int rt){
    MAX[rt]=max(MAX[rt<<1],MAX[rt<<1|1]);
}
void build(int l,int r,int rt){
    MAX[rt]=0;
    if(l==r){
        MAX[rt]=s[l];
        return;
    }
    int m=(l+r)>>1;
    build(lson);
    build(rson);
    pushup(rt);
}

void update(int p,int x,int l,int r,int rt){
    if(l==r){
        MAX[rt]=x;
        return;
    }
    int m=(l+r)>>1;
    if(p<=m)
        update(p,x,lson);
    else
        update(p,x,rson);
    pushup(rt);
}

int query(int L,int R,int l,int r,int rt){
    if(L<=l&&r<=R){
        return MAX[rt];
    }
    int m=(l+r)>>1;
    int res=0;
    if(L<=m){
        res=max(res,query(L,R,lson));
    }
    if(R>m){
        res=max(res,query(L,R,rson));
    }
    return res;
}


int main(){
    int n;
    while(scanf("%d",&n)!=EOF){
        fill(s,s+MAXN,-INF);
        for(int i=1;i<=n;i++){
            scanf("%lld",a+i);
            s[i]=a[i]-i;
        }
        for(int j=1;j<=n;j++)
            scanf("%lld",b+j);
        sort(b+1,b+n+1);
        build(1,2*n,1);
        long long sum=0;
        int tot=1;
        for(int i=n+1;i<=2*n;i++){
            int x=query(b[tot],i-1,1,2*n,1);
            s[i]=x-i;
            sum=(sum+(long long)x+MODE)%MODE;
            if(s[i]>0)
                update(i,s[i],1,2*n,1);
            tot++;
        }
        printf("%lld\n",sum);
    }
}