题目传送门 /* 题意:加上适当的括号,改变计算顺序使得总的计算次数最少 矩阵连乘积问题,DP解决:状态转移方程: dp[i][j] = min (dp[i][k] + dp[k+1][j] + p[i-1] * p[k] * p[j]) (i<=k<j) s[i][j] 记录断开的地方(即加括号的位置),回溯法输出结果 */ #include <cstdio> #include <cstring> #include <string> #include &l…
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1276 裸的矩阵连乘问题. #include<iostream> #include<algorithm> #include<string> #include<cstring> #include<cmath> using namespace std; int n; ]; ][], s[][]; void dp() { ; i…
传送门 •题意 矩阵 A(n×m) 和矩阵 B(m×k) 相乘,共做 n×m×k 次乘法运算: 给你 n 个矩阵,求这 n 个矩阵的最优结合方式,使得做的总乘法运算次数最少: •题解 定义dp(i,j)表示第[i,...j]个矩阵的最小的乘法运算次数: 定义d(i,j)表示第[i,...j]个矩阵的最优解从d(i,j)分开,即(Ai×Ai+1×......×Ad(i,j))×(Ad(i,j)+1×......×Aj): 求解dp(i,j)的时候,求出最优解,并记录是从那个位置分开使得其最优: 输…
/* 题意:矩阵相乘的最少的步数 dp[i][j]=min(dp[i][j], dp[i][k]+dp[k+1][j]+num[i-1]*num[k]*num[j]); 表示的是第i个矩阵到第j个矩阵相乘的最少步数 sign[i][j]表示的是第i个矩阵到第j个矩阵相乘的最少步数是由第i个矩阵到第sign[i][j]个矩阵相乘最少步数 和第sign[i][j]+1个矩阵到第j个矩阵相乘最少步数的得到的最小值! */ #include<iostream> #include<cstring&…
id=19208">题目:Optimal Array Multiplication Sequence 题目大意:给出N个矩阵相乘.求这些矩阵相乘乘法次数最少的顺序. 解题思路:矩阵相乘不满足交换率但满足结合率.dp[i][j] 代表第1个矩阵到第j个矩阵之间的最少的乘法次数,转移状态方程:dp[i][j] = Min(dp[i][k] + dp[k + 1][j]  + A[i - 1] * A[k] *A[j]) k>= i && k <= j - 1.A0A…
C. Sasha and Array time limit per test:5 seconds memory limit per test:256 megabytes input:standard input output: standard output Sasha has an array of integers a1, a2, ..., an. You have to perform m queries. There might be queries of two types: 1 l…
Matrix multiplication                                                                           Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Problem Description Given two matrices A and B of size n×n, find the…
Lucky Coins Sequence Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 608 Accepted Submission(s): 319 Problem Description As we all know,every coin has two sides,with one side facing up and another…
题目链接:hdu_5950_Recursive sequence 题意:递推求解:F(n) = 2*F(n-2) + F(n-1) + n4 和F(1) = a,F(2) = b: 题解: 一看数据范围,肯定矩阵加速递推,不过公式不是线性的,需要把公式转换为线性的公式 #include<bits/stdc++.h> #define F(i,a,b) for(int i=a;i<=b;i++) using namespace std; typedef long long ll; ; ll…
题目链接:传送门 题目: Recursive sequence Time Limit: / MS (Java/Others) Memory Limit: / K (Java/Others) Total Submission(s): Accepted Submission(s): Problem Description Farmer John likes to play mathematics games with his N cows. Recently, they are attracted…