Codeforces Round #485 (Div. 2) C Three displays 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 si<sj<sk should be held.
The rent cost is for the ci. Please determine the smallest cost Maria Stepanovna should pay.
The first line contains a single integer 3≤n≤3000) — the number of displays.
The second line contains 1≤si≤109) — the font sizes on the displays in the order they stand along the road.
The third line contains 1≤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 si<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 40+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 ",ans==INF?-1:ans); 24 return 0; 25 }