矩阵连乘问题

来源：互联网发布：安卓防沉迷软件编辑：程序博客网时间：2024/05/17 02:07

没有输出矩阵连乘的结果，若要输出，加一个函数，本次仅输出从哪断开求矩阵最简

#include <iostream>using namespace std;int m[105][105], s[105][105] ;void MatrixChain(int *p, int n){//p 每个矩阵的行和列, n 一共有多少个矩阵, m 记录最小次数, s 记录断开位置 for(int i = 1; i <= n; ++i) m[i][i] = 0;for(int r = 2; r <= n; ++r){for(int i = 1; i <= n - r + 1; i++){//以上两从循环遍历出了所有情况 1-n 分成两组的所有情况 int j = r + i - 1;m[i][j] = m[i + 1][j] + p[i - 1] * p[i] * p[j];s[i][j] = i;for(int k = i + 1; k < j; ++k){//找最小值 int t = m[i][k] + m[k + 1][j] + p[i - 1] * p[k] * p[j];if(t < m[i][j]){m[i][j] = t;s[i][j] = k;}}}} }void Trackback(int i, int j){//递归输出 if(i == j){return;}Trackback(i, s[i][j]);Trackback(s[i][j] + 1, j);cout << "Multiply A " << i << "," << s[i][j];cout << "and A " << (s[i][j] + 1) << "," << j << endl;}int main(){int n;int p[105];cin >> n;for(int i = 0; i <= n; ++i){cin >> p[i] ;}MatrixChain(p, n);Trackback(1, n);return 0;}/*51 2 3 4 5 6*/

用备忘录的方法，如果m[i][j] 已经求过了，就不在求，直接用

#include <iostream>using namespace std;int m[105][105], s[105][105] ;int p[105]; int LookupChain(int i, int j){if(m[i][j] > 0)return m[i][j];if(i == j)return 0;int u = LookupChain(i, i) + LookupChain(i + 1, j) + p[i - 1] * p[i] * p[j];s[i][j] = i;for(int k = i + 1; k < j; ++k){int t = LookupChain(i, k) + LookupChain(k + 1, j) + p[i - 1] * p[k] * p[j];if(t < u){u = t;s[i][j] = k;} } m[i][j] = u;return u;}int MemorizedMatrixChain(int n){for(int i = 1; i <= n; ++i){for(int j = 1; j <= n; ++j)m[i][j] = 0;}return LookupChain(1, n);} void Trackback(int i, int j){//递归输出 if(i == j){return;}Trackback(i, s[i][j]);Trackback(s[i][j] + 1, j);cout << "Multiply A " << i << "," << s[i][j];cout << "and A " << (s[i][j] + 1) << "," << j << endl;}int main(){int n;cin >> n;for(int i = 0; i <= n; ++i){cin >> p[i] ;}MemorizedMatrixChain(n);Trackback(1, n);return 0;}

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