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  • CF987C Three displays 暴力

    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,nnn,3<=n<=30003<=n<=30003<=n<=3000

    一行nnn个整数,即sss

    再一行nnn个整数,即ccc

    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;
    }
    
    EPFL - Fighting
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  • 原文地址:https://www.cnblogs.com/zxyqzy/p/10273592.html
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