算法分析设计实践——矩阵链乘法

问题：

设N个矩阵序列，其中第i个矩阵式p[i - 1] * p[i]阶矩阵，给定矩阵的向量P，求一种乘法次序，使得基本运算总次数最小

解析：

设A[i][j] 为 ∏^j_{k = i}a[k]

设cnt[i][j]为A[i][j]的最少运算次数

cnt[i][j] = min(cnt[i][k] + cnt[k + 1][j] + p[i - 1] * p[k] * p[j])

设计（核心代码）：

    for (int len = 2; len <= n; ++len)
    {
        for (int i = 1; i + len - 1 <= n; ++i)
        {
            int j = i + len - 1;
            cnt[i][j] = inf;
            s[i][j] = 0;
            for (int k = i; k < j; ++k)
            {
                int tmp = cnt[i][k] + cnt[k + 1][j] + p[i - 1] * p[k] * p[j];
                if (tmp < cnt[i][j])
                {
                    cnt[i][j] = tmp;
                    s[i][j] = k;
                }
            }
        }
    }

分析：

O（n³）

源码：

https://github.com/BambooCertain/Algorithm.git

#include<cstdio>
#include<string.h>
#include<algorithm>
#include<cmath>
#include<iostream>
#include<vector>
#include<queue>
#include<set>
#include<map>
#include<cctype>
#define ios ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define mem(a,x) memset(a,x,sizeof(a))
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid + 1,r
#define P pair<int,int>
#define ull unsigned long long
using namespace std;
typedef long long ll;
const int maxn = 2e5 + 10;
const ll mod = 998244353;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-7;

inline ll read()
{
    ll X = 0, w = 0; char ch = 0;
    while (!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    return w ? -X : X;
}

int cnt[1000][1000] , s[1000][1000];
int p[1000];
int n;
int main()
{
    n = read();
    for (int i = 0; i <= n; ++i)
    {
        p[i] = read();  cnt[i][i] = 0;
    }
    for (int len = 2; len <= n; ++len)
    {
        for (int i = 1; i + len - 1 <= n; ++i)
        {
            int j = i + len - 1;
            cnt[i][j] = inf;
            s[i][j] = 0;
            for (int k = i; k < j; ++k)
            {
                int tmp = cnt[i][k] + cnt[k + 1][j] + p[i - 1] * p[k] * p[j];
                if (tmp < cnt[i][j])
                {
                    cnt[i][j] = tmp;
                    s[i][j] = k;
                }
            }
        }
    }
    cout << cnt[1][n] << endl;
    return 0;
}