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cost=∑i∗ci 最小，归并要求原数组的数的顺序在新数组中不改变。

题解：

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;  //
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //
const int mod = 1e8+;
const double esp = 1e-;
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=;int f=;char ch=getchar();
    while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();
    while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();
    return x=f?-x:x;
}

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

            const int maxn = 1e5+;
            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*];

            void add(int op, int l,int r,int &now){
                    for(int i=l; i<=r; i++){
                        if(op==) C[now++] = A[i];
                        else C[now++] = B[i];
                    }
            }
int main(){
            int n,m;    scanf("%d%d", &n, &m);

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

            int tot2 = ;

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

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

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