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
", ans);
	}
	return 0;
}