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

題意：

出題人有兩個(gè)數(shù)組,A,B，請你把兩個(gè)數(shù)組歸并起來使得$\mathrm{cost=\sum i?ci\; 最小，歸并要求原數(shù)組的數(shù)的順序在新數(shù)組中不改變。}$

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$\mathrm{題解：}$

$\mathrm{先分別處理A和B數(shù)組，把A先分成n段，把某段均值大于前面的一段，就把這兩段合并。處理完A，B段后就可以取大原則歸并。}$

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#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_queuetypedef 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;//這是一個(gè)大根堆q //priority_queue<int,vector<int>,greater<int> >q;//這是一個(gè)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) //用來壓行 #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; //黃金分割點(diǎn) 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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轉(zhuǎn)載于:https://www.cnblogs.com/ckxkexing/p/10294097.html

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