Codeforces Round #485(div 2) C

题目描述

n个显示器排成一排，第i个显示器能显示\(s[i]\)个字，价格为\(c[i]\)，选出三个编号为\(i<j<k\)的显示器，并且满足\(s[i]<s[j]<s[k]\)，求满足条件的最小花费

思路

枚举中间显示器的编号\(j\)，向左找满足\(s[i]<s[j]\)的花费最小的，向右找满足\(s[j]<s[k]\)的花费最小的

代码

#include <bits/stdc++.h>

using namespace std;

const int maxn = 3005;
const long long inf = 0x3f3f3f3f3f3f3f3f;

typedef long long ll;

ll s[maxn];
ll c[maxn];
ll n;
int main(){
    ll c1,c2,c3;
    ll ans;
    scanf("%lld", &n);
    for (int i = 0; i < n; ++i){
        scanf("%lld", s+i);
    }
    for (int i = 0; i < n; ++i){
        scanf("%lld", c+i);
    }
    ans = inf;
    // 枚举中间显示器的编号j
    for (int j = 1; j < n - 1; ++j){
        c2 = c[j];
        c1 = inf;
        c3 = inf;
        // 向左找满足s[i]<s[j]的花费最小的
        for (int i = 0; i < j; ++i){
            if (s[i] < s[j] && c[i] < c1){
                c1 = c[i];
            }
        }
        // 向右找满足s[j]<s[k]的花费最小的
        for (int k = j + 1; k < n; ++k){
            if (s[k] > s[j] && c[k] < c3){
                c3 = c[k];
            }
        }
        if (c1 == inf || c3 == inf) continue;
        ans = min(ans, c1 + c2 + c3);
    }
    if (ans == inf){
        puts("-1");
    }
    else{
        printf("%lld\n", ans);
    }
    return 0;
}