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.
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.
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.
5 2 4 5 4 10 40 30 20 10 40
90
3 100 101 100 2 4 5
-1
10 1 2 3 4 5 6 7 8 9 10 10 13 11 14 15 12 13 13 18 13
33
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.
思路:思维题,因为要选三个城市,我们可以预处理出满足条件的两个城市之间的最小值,在枚举第三个城市即可。
过题代码:
#include <bits/stdc++.h>
using namespace std;
const int maxn=3010,inf=0x3f3f3f3f;
int s[maxn],c[maxn],f[maxn],dp[maxn];
int n,ans;
int main(){
scanf("%d",&n);
for (int i=1;i<=n;i++) scanf("%d",&s[i]);
for (int i=1;i<=n;i++) scanf("%d",&c[i]);
dp[1]=inf;
for (int i=2;i<=n;i++){
dp[i]=inf;
for (int j=i-1;j>=1;j--){
if(s[i]>s[j])
dp[i]=min(dp[i],c[i]+c[j]);
}
// printf("%d ",dp[i]);
}
// printf("
");
ans=inf;
for (int i=3;i<=n;i++){
f[i]=inf;
for (int j=i-1;j>=1;j--){
if(s[j]<s[i]) f[i]=min(f[i],c[i]+dp[j]);
}
// printf("%d ",f[i]);
ans=min(ans,f[i]);
}
// printf("
");
if(ans==inf) puts("-1");
else printf("%d
",ans);
return 0;
}