zoukankan      html  css  js  c++  java
  • Xenia and Colorful Gems CodeForces 1337D[binary search +data structures implementation+ math sortings]

    Xenia is a girl being born a noble. Due to the inflexibility and harshness of her family, Xenia has to find some ways to amuse herself.

    snoop dooogg~~~ 

    Recently Xenia has bought nrnr red gems, ngng green gems and nbnb blue gems. Each of the gems has a weight.

    Now, she is going to pick three gems.

    Xenia loves colorful things, so she will pick exactly one gem of each color.

    Xenia loves balance, so she will try to pick gems with little difference in weight.

    Specifically, supposing the weights of the picked gems are xx, yy and zz, Xenia wants to find the minimum value of (xy)2+(yz)2+(zx)2(x−y)2+(y−z)2+(z−x)2. As her dear friend, can you help her?

    Input

    The first line contains a single integer tt (1t1001≤t≤100)  — the number of test cases. Then tt test cases follow.

    The first line of each test case contains three integers nr,ng,nbnr,ng,nb (1nr,ng,nb1051≤nr,ng,nb≤105)  — the number of red gems, green gems and blue gems respectively.

    The second line of each test case contains nrnr integers r1,r2,,rnrr1,r2,…,rnr (1ri1091≤ri≤109)  — riri is the weight of the ii-th red gem.

    The third line of each test case contains ngng integers g1,g2,,gngg1,g2,…,gng (1gi1091≤gi≤109)  — gigi is the weight of the ii-th green gem.

    The fourth line of each test case contains nbnb integers b1,b2,,bnbb1,b2,…,bnb (1bi1091≤bi≤109)  — bibi is the weight of the ii-th blue gem.

    It is guaranteed that nr105∑nr≤105, ng105∑ng≤105, nb105∑nb≤105 (the sum for all test cases).

    Output

    For each test case, print a line contains one integer  — the minimum value which Xenia wants to find.

    Example

    Input
    5
    2 2 3
    7 8
    6 3
    3 1 4
    1 1 1
    1
    1
    1000000000
    2 2 2
    1 2
    5 4
    6 7
    2 2 2
    1 2
    3 4
    6 7
    3 4 1
    3 2 1
    7 3 3 4
    6
    
    Output
    14
    1999999996000000002
    24
    24
    14
    

    Note

    In the first test case, Xenia has the following gems:

    If she picks the red gem with weight 7, the green gem with weight 6, and the blue gem with weight 4, she will achieve the most balanced selection with (xy)2+(yz)2+(zx)2=(76)2+(64)2+(47)2=14(x−y)2+(y−z)2+(z−x)2=(7−6)2+(6−4)2+(4−7)2=14.

    ACcode#1

    #include <bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    const ll inf = 4e18+10;
    const int mod = 1000000007;
    const int mx = 200010; //check the limits, dummy
    typedef pair<int, int> pa;
    const double PI = acos(-1);
    ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
    #define swa(a,b) a^=b^=a^=b
    #define re(i,a,b) for(int i=(a),_=(b);i<_;i++)
    #define rb(i,a,b) for(int i=(b),_=(a);i>=_;i--)
    #define clr(a) memset(a, 0, sizeof(a))
    #define lowbit(x) ((x)&(x-1))
    #define mkp make_pair
    void sc(int& x) { scanf("%d", &x); }void sc(int64_t& x) { scanf("%lld", &x); }void sc(double& x) { scanf("%lf", &x); }void sc(char& x) { scanf(" %c", &x); }void sc(char* x) { scanf("%s", x); }
    ll  m, n,t,y,z,k,sum=0,ans=0;
    ll fuck(ll x, ll y, ll z) {
        return (x - y) * (x - y) + (y - z) * (y - z) + (x - z) * (x - z);
    }
    int main()
    {
        ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
        cin >> t;
        while (t--)
        {
            ll candy[3];
            vector<ll>v[3];
            cin >> candy[0] >> candy[1] >> candy[2];
            re(p, 0, 3) {
                v[p].resize(candy[p]);
                re(i, 0, candy[p])cin >> v[p][i];
                sort(v[p].begin(), v[p].end());
            }
            ans = inf;
            re(p1, 0, 3) {
                re(p2, 0, 3) {
                    re(p3, 0, 3) {
                        if (p1 == p2 || p2 == p3 || p1 == p3)continue;
                        for (auto x : v[p1]) {
                            auto i1 = lower_bound(v[p2].begin(), v[p2].end(), x);
                            auto i2 = upper_bound(v[p3].begin(), v[p3].end(), x);
                            if (i1 != v[p2].end() && i2 != v[p3].begin()) {
                                y = *i1;
                                i2--;
                                z = *i2;
                                ans = min(ans, fuck(x, y, z));
                            }
                        }
                    }
                }
            }
            cout << ans << endl;
        }
        return 0;
    }

     ACcode#2

    #include<bits/stdc++.h>
    using namespace std;
    int read() {
        char c=getchar();while(!isdigit(c)) c=getchar();
        int num=0;while(isdigit(c)) num=num*10+c-'0',c=getchar();
        return num;
    }
    inline long long func(int a, int b, int c) {
        return 1ll*(a-b)*(a-b)+1ll*(b-c)*(b-c)+1ll*(c-a)*(c-a);
    }
    inline long long solve(int a[], int b[], int c[], int n1, int n2, int n3) {
        long long ans = 5000000000000000000;
        int p = 0, q = 0;
        for (int i = 1; i <= n1; i++) {
            while (q <= n2 && b[q] < a[i]) ++q;
            while (p < n3 && c[p+1] <= a[i]) ++p;
            if (p && q <= n2) ans = min(ans, func(a[i], b[q], c[p]));
        }
        return ans;
    }
    int a[100001], b[100001], c[100001];
    int main() {
        int t = read();
        while (t--) {
            int n1, n2, n3;
            n1 = read(), n2 = read(), n3 = read();
            for (int i = 1; i <= n1; i++) a[i] = read();
            for (int i = 1; i <= n2; i++) b[i] = read();
            for (int i = 1; i <= n3; i++) c[i] = read();
            sort(a + 1, a + n1 + 1);
            sort(b + 1, b + n2 + 1);
            sort(c + 1, c + n3 + 1);
            long long ans = 5000000000000000000;
            ans = min(ans, solve(a, b, c, n1, n2, n3));
            ans = min(ans, solve(a, c, b, n1, n3, n2));
            ans = min(ans, solve(b, a, c, n2, n1, n3));
            ans = min(ans, solve(b, c, a, n2, n3, n1));
            ans = min(ans, solve(c, a, b, n3, n1, n2));
            ans = min(ans, solve(c, b, a, n3, n2, n1));
            cout << ans << endl;
        }
    }
  • 相关阅读:
    cgroup中对cpu资源控制的方式
    kube-proxy 原理
    k8s集群优化
    k8s集群从一千节点增加到五千台节点遇到的瓶颈
    k8s的service和ep是如何关联和相互影响的
    一个经典pod完整生命周期
    k8s中的pod内几个容器之间的关系是什么
    kubernetes包含几个组件。各个组件的功能是什么。组件之间是如何交互的。
    k8s的pause容器有什么用?是否可以去掉?
    强制删除pod,namespace等
  • 原文地址:https://www.cnblogs.com/xxxsans/p/12712419.html
Copyright © 2011-2022 走看看