CF987C Three displays 暴力

最新推荐文章于 2021-01-24 01:41:28 发布

weixin_33739541

最新推荐文章于 2021-01-24 01:41:28 发布

阅读量96

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/zxyqzy/p/10273592.html

本文解析一道算法题，目标是在限定条件下找出三个元素的组合，使这三个元素的特定属性之和达到最小。文章详细介绍了问题背景、输入输出格式及解题思路，采用枚举法结合暴力搜索策略，最终实现O(N^2)的时间复杂度。

题意翻译

题目大意：

$nnn个位置，每个位置有两个属性s,cs,cs,c，要求选择3个位置i,j,ki,j,ki,j,k，使得si<sj<sks_i<s_j<s_ksi<sj<sk，并使得ci+cj+ckc_i+c_j+c_kci+cj+ck最小$

输入格式：

一行一个整数，

一行

再一行

输出格式：

输出一个整数，即最小的c_i+c_j+c_k

枚举 j ，分段暴力；

O(N^2)；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 900005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-4
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
    ll x = 0;
    char c = getchar();
    bool f = false;
    while (!isdigit(c)) {
        if (c == '-') f = true;
        c = getchar();
    }
    while (isdigit(c)) {
        x = (x << 1) + (x << 3) + (c ^ 48);
        c = getchar();
    }
    return f ? -x : x;
}

ll gcd(ll a, ll b) {
    return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }


/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
    if (!b) {
        x = 1; y = 0; return a;
    }
    ans = exgcd(b, a%b, x, y);
    ll t = x; x = y; y = t - a / b * y;
    return ans;
}
*/

int n;
ll s[3006];
ll c[3006];
ll minn1[4000];
ll minn2[4000];

int main() {
    //ios::sync_with_stdio(0);
    rdint(n);
    for (int i = 1; i <= n; i++)rdllt(s[i]);
    for (int i = 1; i <= n; i++)rdllt(c[i]);
    ll ans = inf;
    c[0] = c[n + 1] = inf;
    for (int i = 2; i < n; i++) {
        int l = 0, r = n + 1;
        for (int j = 1; j < i; j++) {
            if (s[j] < s[i])
                if (c[j] < c[l])l = j;
        }
        for (int j = i + 1; j <= n; j++) {
            if (s[j] > s[i])
                if (c[j] < c[r])r = j;
        }
        if (l != 0 && r != n + 1)ans = min(ans, c[i] + c[l] + c[r]);
    }
    if (ans == inf)cout << -1 << endl;
    else cout << ans << endl;
    return 0;
}

转载于:https://www.cnblogs.com/zxyqzy/p/10273592.html