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 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
Copy
52 4 5 4 1040 30 20 10 40
output
Copy
90
input
Copy
3100 101 1002 4 5
output
Copy
-1
input
Copy
101 2 3 4 5 6 7 8 9 1010 13 11 14 15 12 13 13 18 13
output
Copy
33
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.
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。
1 #include <bits/stdc++.h>
2 #define INF 0x3f3f3f3f
3 using namespace std;
4 const int N = 3010;
5 int n, s[N], c[N];
6 int dp[N][3];
7 int main() {
8 cin >> n;
9 memset(dp, INF, sizeof(dp));
10 for(int i = 0; i < n; i ++) cin >> s[i];
11 for(int i = 0; i < n; i ++) cin >> c[i], dp[i][0] = c[i];
12 for(int i = 1; i < 3; i ++) {
13 for(int j = 0; j < n; j ++) {
14 for(int k = 0; k < j; k ++) {
15 if(s[j] > s[k]) {
16 dp[j][i] = min(dp[j][i],dp[k][i-1] + c[j]);
17 }
18 }
19 }
20 }
21 int ans = INF;
22 for(int i = 0; i < n; i ++) ans = min(dp[i][2], ans);
23 printf("%d\n",ans==INF?-1:ans);
24 return 0;
25 }
原文地址：https://www.cnblogs.com/xingkongyihao/p/9113880.html