Codeforces Round #485 (Div. 2) C Three displays

C. Three displays

It is the middle of 2018 and Maria Stepanovna, who lives outside Krasnokamensk (a town in Zabaikalsky region), wants to rent three displays to highlight an important problem.

There are $\mathrm{nn displays\; placed\; along\; a\; road,\; and\; the ii-th\; of\; them\; can\; display\; a\; text\; with\; font\; size sisi only.\; Maria\; Stepanovna\; wants\; to\; rent\; such\; three\; displays\; with\; indices i<j<ki<j<k that\; the\; font\; size\; increases\; if\; you\; move\; along\; the\; road\; in\; a\; particular\; direction.\; Namely,\; the\; condition si<sj<sksi<sj<sk should\; be\; held.}$

The rent cost is for the $\mathrm{ii-th\; display\; is cici.\; Please\; determine\; the\; smallest\; cost\; Maria\; Stepanovna\; should\; pay.}$

Input

The first line contains a single integer $\mathrm{nn (3\le n\le 30003\le n\le 3000) —\; the\; number\; of\; displays.}$

The second line contains $\mathrm{nn integers s1,s2,\dots ,sns1,s2,\dots ,sn (1\le si\le 1091\le si\le 109) —\; the\; font\; sizes\; on\; the\; displays\; in\; the\; order\; they\; stand\; along\; the\; road.}$

The third line contains $\mathrm{nn integers c1,c2,\dots ,cnc1,c2,\dots ,cn (1\le ci\le 1081\le ci\le 108) —\; the\; rent\; costs\; for\; each\; display.}$

Output

If there are no three displays that satisfy the criteria, print -1. Otherwise print a single integer — the minimum total rent cost of three displays with indices $\mathrm{i<j<ki<j<k such\; that si<sj<sksi<sj<sk.}$

Examples

input

Copy

5
2 4 5 4 10
40 30 20 10 40

output

Copy

90

input

Copy

3
100 101 100
2 4 5

output

Copy

-1

input

Copy

10
1 2 3 4 5 6 7 8 9 10
10 13 11 14 15 12 13 13 18 13

output

Copy

33

Note

In the first example you can, for example, choose displays $\mathrm{11, 44 and 55,\; because s1<s4<s5s1<s4<s5 (2<4<102<4<10),\; and\; the\; rent\; cost\; is 40+10+40=9040+10+40=90.}$

In the second example you can't select a valid triple of indices, so the answer is -1.

取三个数，使的si < sj < sk  并且使得ci + cj + sk 最小。

dp做法：

dp[i][j] 表示选取了 j 个是并且第 si 是最大时，获取的值最小是多少。

还有一种方法。循环到第 i 个数时，求左边比si小的最小的 c 数，右边比si 大的最小的 c 数。

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
const int N = 3010;
int n, s[N], c[N];
int dp[N][3];
int main() {
    cin >> n;
    memset(dp, INF, sizeof(dp));
    for(int i = 0; i < n; i ++) cin >> s[i];
    for(int i = 0; i < n; i ++) cin >> c[i], dp[i][0] = c[i];
    for(int i = 1; i < 3; i ++) {
        for(int j = 0; j < n; j ++) {
            for(int k = 0; k < j; k ++) {
                if(s[j] > s[k]) {
                    dp[j][i] = min(dp[j][i],dp[k][i-1] + c[j]);
                }
            }
        }
    }
    int ans = INF;
    for(int i = 0; i < n; i ++) ans = min(dp[i][2], ans);
    printf("%d
",ans==INF?-1:ans);
    return 0;
}