【CodeForces 987C】Three displays

【链接】 我是链接,点我呀:)

【题意】

【题解】

动态规划
设dp[i][j]表示前i个数字,选了j个的最小花费。
dp[i][j] = min(dp[k][j-1]+b[i]);//其中a[i]>a[k]且ka[k]的位置k才有可能。
复杂度O(N^2)

【代码】

#include <bits/stdc++.h>

using namespace std;

const int N = 3e3;
const int INF = 4e8;

int n;
int a[N+10],b[N+10];
int dp[N+10][4];

int main()
{
    #ifdef LOCAL_DEFINE
        freopen("rush.txt","r",stdin);
    #endif // LOCAL_DEFINE
    ios::sync_with_stdio(0),cin.tie(0);
    cin >> n;
    for (int i = 1;i <= n;i++) cin >> a[i];
    for (int i = 1;i <= n;i++) cin >> b[i];
    for (int i = 1;i <= 3;i++)
        for (int j = 1;j <= n;j++)
            dp[j][i] = INF;
    for (int i = 1;i <= n;i++) dp[i][1] = b[i];

    for (int j = 2;j <= 3;j++)
        for (int i = j;i <= n;i++)
            for (int k = j-1;k <= i-1;k++)
                if (a[i]>a[k] && dp[i][j]>dp[k][j-1]+b[i]){
                    dp[i][j] = dp[k][j-1]+b[i];
                }
    int ans = INF;
    for (int i = 3;i <= n;i++)
        ans = min(ans,dp[i][3]);
    if (ans==INF)
        cout<<-1<<endl;
    else
        cout<<ans<<endl;

    return 0;
}