算法分析第八次作业

1. 问题

  给定n个矩阵｛A1,A2,…,An｝，其中Ai与Ai+1是可乘的，i=1,2 ,…,n-1。如何确定计算矩阵连乘积的计算次序，使得依此次序计算矩阵连乘积需要的数乘次数最少。

2. 解析

     这其实是个很简单的问题，区间dp模板题，dp[i][j]，表示区间i到j合并的最小值，枚举k表示分割点，dp[1][n]即为答案

 转移方程：dp[i][j]=min(dp[i][j],dp[i][k]+d[k+1][j]+b[i]*b[k+1]*b[j+1]

3. 设计

  

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for (int h = 1; h < n; h++) {//区间dp     for (int i = 1; i <= n - h; i++) {         int j = i + h;         for (int k = i; k <= j - 1; k++) {             if(dp[i][j]==0) dp[i][j]=dp[i][k] + dp[k + 1][j] + b[i] * b[k + 1] * b[j + 1];             else dp[i][j]=min(dp[i][j],dp[i][k] + dp[k + 1][j] + b[i] * b[k + 1] * b[j + 1]);         }     } }

 

 

4. 分析

 复杂度 O(n3)

5. 源码

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

+ View Code?

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#include<stdio.h> #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 = 600 + 11; const int maxm = 600 + 11; const int mod = 1e9 + 7; const double eps = 1e-5; 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 int b[maxn], dp[maxn][maxn],a[maxn]; int main() {     int n;     memset(dp, 0, sizeof dp);     cin >> n;     cin >> b[1] >> b[2];     for (int i = 3; i <= n + 1; i++) {         cin >> a[i] >> b[i];     }     for (int h = 1; h < n; h++) {//区间dp         for (int i = 1; i <= n - h; i++) {             int j = i + h;             for (int k = i; k <= j - 1; k++) {                 if(dp[i][j]==0) dp[i][j]=dp[i][k] + dp[k + 1][j] + b[i] * b[k + 1] * b[j + 1];                 else dp[i][j]=min(dp[i][j],dp[i][k] + dp[k + 1][j] + b[i] * b[k + 1] * b[j + 1]);             }         }     }     cout << dp[1][n] << endl;     return 0; }