【CodeForces - 987C 】Three displays （dp，最长上升子序列类问题，三元组问题）

题干：

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 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 ii-th display is cici. Please determine the smallest cost Maria Stepanovna should pay.

Input

The first line contains a single integer nn (3≤n≤30003≤n≤3000) — the number of displays.

The second line contains nn integers s1,s2,…,sns1,s2,…,sn (1≤si≤1091≤si≤109) — the font sizes on the displays in the order they stand along the road.

The third line contains nn integers c1,c2,…,cnc1,c2,…,cn (1≤ci≤1081≤ci≤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 i<j<ki<j<k such that si<sj<sksi<sj<sk.

Examples

Input

5
2 4 5 4 10
40 30 20 10 40

Output

Input

3
100 101 100
2 4 5

Output

-1

Input

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

Output

Note

In the first example you can, for example, choose displays 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.

题目大意：

解题报告：

三元组问题。

类似最长上升子序列的做法，但是维护的不是长度，而是的最小值。

AC代码：

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int MAX = 3000 + 5;
const ll INF = 0x3f3f3f3f3f3f3f3f;
ll a[MAX];
ll b[MAX];
ll dp[MAX];
int main()
{int n;ll ans = INF;cin>>n;for(int i = 1; i<=n; i++) {scanf("%lld",&a[i]);}for(int i = 1; i<=n; i++) {scanf("%lld",&b[i]);}memset(dp,INF, sizeof(dp));for(int i =2; i<=n-1; i++) {//其实不用到n，因为还要留一个给第三个display，所以这里递归到n-1.for(int j = 1; j<i; j++) {if(a[i] > a[j])dp[i] = min(dp[i], b[i] + b[j]);}}
//    printf("%lld\n",dp[3]);for(int i = 1; i<=n-1; i++) {for(int j = i+1; j<=n; j++) {if(a[j] > a[i])ans = min(ans ,dp[i]+b[j]);
//            printf("%lld %lld ==== %d %d\n",dp[i],b[j] ,i,j);}}if(ans == INF) printf("-1\n");else cout << ans<<endl;return 0 ;
}

或者直接dp解：

详见博客https://blog.csdn.net/xs18952904/article/details/80504296

20191003

忽然发现今天随便刷codeforce的C题又刷到这道题了，特此更新：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
ll dp1[MAX],dp2[MAX],a[MAX],c[MAX];
int n;
int main()
{cin>>n;for(int i = 1; i<=n; i++) scanf("%lld",a+i),dp1[i]=1e18;for(int i = 1; i<=n; i++) scanf("%lld",c+i),dp2[i]=1e18;for(int i = 1; i<=n; i++) {for(int j = 1; j<i; j++) {if(a[i] > a[j]) dp1[i] = min(dp1[i],c[j]+c[i]);}}for(int i = 1; i<=n; i++) {for(int j = 1; j<i; j++) {if(a[i] > a[j]) dp2[i] = min(dp2[i],dp1[j]+c[i]);}}ll ans = *min_element(dp2+1,dp2+n+1);if(ans == (ll)1e18) ans = -1;printf("%lld\n",ans);return 0 ;
}

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