[HDU]-6047 Maximum Sequence

Maximum Sequence

URL: http://acm.hdu.edu.cn/showproblem.php?pid=6047

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 71 Accepted Submission(s): 32

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

其实还有 o(n)的解法

#include<stdio.h>#include<string.h>#include<queue>#include<algorithm>using namespace std;typedef pair<int,int> P;const int MAXN = 250000 + 5;int b, a[MAXN], num[MAXN];const int MOD = 1e9 + 7;int main(){    int n;    while(scanf("%d", &n) != EOF) {        priority_queue<P, vector<P>, less<P> > que;        for(int i = 1; i <= n; ++i){            scanf("%d", a + i);            a[i] -= i;            que.push(P(a[i], i));        }        for(int i = 1; i <= n; ++i){            scanf("%d", &b);            ++num[b];        }        int ans = 0, tol = n + 1;        for(int i = 1; i <= n; ++i){            if(num[i] == 0) continue ;            while(que.top().second < i) que.pop();            P p = que.top();            int val = p.first, id = p.second;        //  printf("%2d -> %2d\n", i, val);            ans += 1LL * val * num[i];            ans %= MOD;            for(int j = 0; j < num[i]; ++j) {                que.push(P(val - tol, tol));                ++tol;            }            num[i] = 0;        }        printf("%d\n", ans);    }    return 0;}