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。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#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; }
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