Codeforces-987C - Three displays - 动态规划

                                        转自： https://blog.youkuaiyun.com/xs18952904/article/details/80504296

真的还有很长的路要走啊！

题目

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 n displays placed along a road, and the i-th of them can display a text with font size si only. Maria Stepanovna wants to rent such three displays with indices i < j < k that the font size increases if you move along the road in a particular direction. Namely, the condition si < sj < sk should be held.

The rent cost is for the i-th display is ci. Please determine the smallest cost Maria Stepanovna should pay.

Input

The first line contains a single integer n (3 ≤ n ≤ 3 000) — the number of displays.

The second line contains n integers s1 , s2 , … , sn (1 ≤ si ≤ 109 ) — the font sizes on the displays in the order they stand along the road.

The third line contains n integers c1 , c2 , … , cn (1 ≤ 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 < k such that si < sj < sk.

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题意

  给你一个长度为n的序列，每个序列有一个权值sisi和花费cici，让你从中选出3个值，使得i<j<ki<j<k且si<sj<sksi<sj<sk，如果存在输出最小话费，不存在的话输出-1。

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思路

  因为只用选出三个物品，f[i,k]f[i,k]代表第ii个物品作为选出的三个物品中的第kk个物品的最小花费，显然有：f[i,k]=min(f[j,k−1]+c[i])f[i,k]=min(f[j,k−1]+c[i])，其中：j<ij<i且s[j]<s[i]s[j]<s[i]。

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实现

#include <bits/stdc++.h>
using namespace std;
const int maxn = 3007, inf = 0x3f3f3f3f;
int f[maxn][4], n, val[maxn], cost[maxn], ans = inf;
int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", val + i);
    for (int i = 1; i <= n; i++) scanf("%d", cost + i);
    memset(f, 0x3f, sizeof(f));
    for (int i = 1; i <= n; i++) {
        f[i][1] = cost[i];
        for (int k = 2; k <= 3; k++)
            for (int j = 1; j < i; j++)
                if (val[j] < val[i]) f[i][k] = std::min(f[i][k], f[j][k - 1] + cost[i]);
    }
    for (int i = 1; i <= n; i++) ans = std::min(ans, f[i][3]);
    printf("%d\n", ans == inf ? -1 : ans);
    return 0;

}