C. Three displays（简单dp）

C. Three displays（简单dp）

C. Three displays

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

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

Examples

Copy

5 2 4 5 4 10 40 30 20 10 40

Copy

90

Copy

3 100 101 100 2 4 5

Copy

-1

Copy

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

Copy

33

题目大概：

给出n个数，每个数都有价值。找出价值最小的三元组序列，满足数 i

思路：

dp一下。

dp【i】【j】是以i结尾的 j 元组的最小价值。然后状态转移一下就行了。

dp【i】【2】=min（dp【i】【2】，dp【k】【1】+w【i】）（1<=k

dp【i】【3】=min（dp【i】【3】，dp【k】【2】+w【i】）（1<=k

代码：

#include using namespace std;const int maxn=3100;const int INF=0x3f3f3f3f;int a[maxn],w[maxn];int dp[maxn][4];int main(){ int n; scanf("%d",&n); for(int i=1;i<=n;i++) { scanf("%d",&a[i]); } for(int i=1;i<=n;i++) { scanf("%d",&w[i]); dp[i][1]=w[i]; } for(int i=1;i<=n;i++) { dp[i][2]=INF; dp[i][3]=INF; for(int j=1;ja[j]) { dp[i][2]=min(dp[j][1]+w[i],dp[i][2]); dp[i][3]=min(dp[j][2]+w[i],dp[i][3]); } } } int sum=INF; for(int i=3;i<=n;i++) { if(dp[i][3]==INF)continue; sum=min(sum,dp[i][3]); } if(sum==INF)printf("-1\n"); else printf("%d\n",sum); return 0;}

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