实验8矩阵链乘法

问题：

设$n$个矩阵序列，其中第i个矩阵是$p[i-1]*p[i]$阶矩阵，给定矩阵链的向量$P$，求一种乘法次序，使得基本运算的总次数最小。

解析：

设$A[i][j]$为$prod_{k=i}^{j}{a[k]}$

设$F[i][j]$为$A[i][j]$的最少运算次数。

$F[i][j]=min(f[i][k]+f[i+1][j]+p[i-1]*p[k]*p[j])$

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;
        f[i][j] = inf;
        s[i][j] = 0;
        for (int k = i; k < j; ++k) {
            int res = f[i][k] + f[k + 1][j] + p[i - 1] * p[k] * p[j];
            if (res < f[i][j]) {
                f[i][j] = res;
                s[i][j] = k;
            }
        }
    }
}

分析：

复杂度：$O(n^3)$

源码：

https://github.com/Big-Kelly/Algorithm

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int inf = 2e9 + 7;
const ll Inf = 1e18 + 7;
const int maxn = 3e3 + 5;
const int mod = 1e9 + 7;

int f[maxn][maxn], s[maxn][maxn];
int p[maxn];
int n;

int main() {
    scanf("%d", &n);
    for (i = 0; i <= n; ++i)    scanf("%d", &p[i]), f[i][i] = 0;
    for (int len = 2; len <= n; ++len) {
        for (int i = 1; i + len - 1 <= n; ++i) {
            int j = i + len - 1;
            f[i][j] = inf;
            s[i][j] = 0;
            for (int k = i; k < j; ++k) {
                int res = f[i][k] + f[k + 1][j] + p[i - 1] * p[k] * p[j];
                if (res < f[i][j]) {
                    f[i][j] = res;
                    s[i][j] = k;
                }
            }
        }
    }
    printf("%d
", f[1][n]);
}