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<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
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
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<j <k（不能打乱给的顺序）。
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<i）
dp【i】【3】=min（dp【i】【3】，dp【k】【2】+w【i】）（1<=k<i）
代码：
#include <bits/stdc++.h>
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;j<i;j++)
{
if(a[i]>a[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;
}
