牛客练习赛38 E 出题人的数组 2018ccpc桂林A题 贪心

https://ac.nowcoder.com/acm/contest/358/E

题意：

出题人有两个数组,A,B，请你把两个数组归并起来使得$\mathrm{cost=\sum i\ast ci\; 最小，归并要求原数组的数的顺序在新数组中不改变。}$

$\mathrm{题解：}$

$\mathrm{先分别处理A和B数组，把A先分成n段，把某段均值大于前面的一段，就把这两段合并。处理完A，B段后就可以取大原则归并。}$

#include <algorithm>
#include <iterator>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <stack>
#include <cmath>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <cassert>

using namespace std;
#define lson (l, mid, rt << 1)
#define rson (mid + 1, r, rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue

typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
typedef pair<int, pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'

#define OKC                      \
    ios::sync_with_stdio(false); \
    cin.tie(0)
#define FT(A, B, C) for (int A = B; A <= C; ++A) //用来压行
#define REP(i, j, k) for (int i = j; i < k; ++i)
#define max3(a, b, c) max(max(a, b), c);
#define min3(a, b, c) min(min(a, b), c);
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e8 + 7;
const double esp = 1e-8;
const double PI = acos(-1.0);
const double PHI = 0.61803399; //黄金分割点
const double tPHI = 0.38196601;

template <typename T>
inline T read(T &x) {
    x = 0;
    int f = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
    return x = f ? -x : x;
}

/*-----------------------showtime----------------------*/

const int maxn = 1e5 + 9;
int A[maxn], B[maxn];

struct node {
    ll sum;
    int cnt;

    friend bool operator<(node a, node b) {
        return a.sum * b.cnt < b.sum * a.cnt;
    }
    friend node operator+(node a, node b) {
        return (node){a.sum + b.sum, a.cnt + b.cnt};
    }
} S[maxn], T[maxn];
int C[maxn * 2];

void add(int op, int l, int r, int &now) {
    for (int i = l; i <= r; i++) {
        if (op == 1)
            C[now++] = A[i];
        else
            C[now++] = B[i];
    }
}

int main() {
    int n, m;
    scanf("%d%d", &n, &m);

    for (int i = 1; i <= n; i++) {
        scanf("%d", &A[i]);
    }
    for (int i = 1; i <= m; i++) {
        scanf("%d", &B[i]);
    }
    int tot1 = 0;
    for (int i = 1; i <= n; i++) {
        S[++tot1] = (node){A[i], 1};
        while (tot1 > 1 && S[tot1 - 1] < S[tot1]) {
            S[tot1 - 1] = S[tot1 - 1] + S[tot1];
            tot1--;
        }
    }

    int tot2 = 0;

    for (int i = 1; i <= m; i++) {
        T[++tot2] = (node){B[i], 1};
        while (tot2 > 1 && T[tot2 - 1] < T[tot2]) {
            T[tot2 - 1] = T[tot2 - 1] + T[tot2];
            tot2--;
        }
    }

    S[++tot1] = (node){-1, 1};
    T[++tot2] = (node){-1, 1};

    int L = 1, R = 1;
    int prel = 1, prer = 1, now = 1;
    while (L < tot1 || R < tot2) {
        if (S[L] < T[R]) {
            add(2, prer, prer + T[R].cnt - 1, now);
            prer += T[R].cnt;
            R++;
        } else {
            add(1, prel, prel + S[L].cnt - 1, now);
            prel += S[L].cnt;
            L++;
        }
    }
    ll ans = 0;
    for (int i = 1; i <= n + m; i++) {
        ans += 1ll * i * C[i];
        // cout<<C[i]<<" ";
    }
    // cout<<endl;
    printf("%lld\n", ans);
    return 0;
}