算法分析第十一次作业

1. 问题

   给出n个数，求出最优前缀码

2. 解析

  每次选出两个数，将两个数合并成一个数，查剩下的有序数组，每一次减少一个数，最后的数就是答案

3. 设计

  我们可以使用权值线段树,来实现O(logn）的插入以及O(logn)的删除，当然我们也可以使用链表等手段，这里我使用了multiset的c++容器来实现插入与删除；

    

for(int i=1;i<=n-1;i++){
    int x=*s.begin();
    s.erase(s.begin());
    int y=*s.begin();
    s.erase(s.begin());
    s.insert(x+y);
}

 

 

4. 分析

  O(nlogn)

 

 

 

5. 源码

https://github.com/Tinkerllt/algorithm-work.git

#include<stdio.h>
#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<bitset>
#include<set>
#include<deque>
#include<queue>
#include<vector>
//#include<unordered_map>
#include<map>
#include<stack>
using namespace std;
#define ll long long
#define ull unsigned long long
#define pii pair<int,int>
#define Pii pair<ll,int>
#define m_p make_pair
#define l_b lower_bound
#define u_b upper_bound
const int inf = 0x3f3f3f3f;
const ll linf = 0x3f3f3f3f3f3f3f3f;
const int maxn = 2e6 + 11;
const int maxm = 600 + 11;
const int mod = 1e9 + 7;
const double eps = 1e-5;
inline ll rd() { ll x = 0, f = 1; char ch = getchar(); while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }while (ch >= '0'&&ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }return x * f; }
inline ll qpow(ll a, ll b, ll p) { ll res = 1; while (b) { if (b & 1) { res *= a; res %= p; }b >>= 1; a = a * a%p; }return res; }
inline ll gcd(ll a, ll b) { if (b == 0) return a; return gcd(b, a%b); }
//iterator
//head
//priority_queue
multiset<int>s; 
int main() {
    int n=rd();
    for(int i=1;i<=n;i++){
        int x=rd();
        s.insert(x);
    } 
    for(int i=1;i<=n-1;i++){
        int x=*s.begin();
        s.erase(s.begin());
        int y=*s.begin();
        s.erase(s.begin());
        s.insert(x+y);
    }
    printf("%d\n",*s.begin());
}