题目链接:https://vjudge.net/problem/POJ-1651 Multiplication Puzzle Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11239   Accepted: 6980 Description The multiplication puzzle is played with a row of cards, each containing a single positive…

Multiplication Puzzle Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7252   Accepted: 4478 Description The multiplication puzzle is played with a row of cards, each containing a single positive integer. During the move player takes one…

//Accepted 200 KB 0 ms //dp区间 //dp[i][j]=min(dp[i][k]+dp[k][j]+a[i]*a[k]*a[j]) i<k<j #include <cstdio> #include <cstring> #include <iostream> using namespace std; ; ; int dp[imax_n][imax_n]; int a[imax_n]; int n; int min(int a,int…

Description The multiplication puzzle is played with a row of cards, each containing a single positive integer. During the move player takes one card out of the row and scores the number of points equal to the product of the number on the card taken…

题目描述: 有若干个矩阵{Ai},元素都为整数且已知矩阵大小. 如果要计算所有矩阵的乘积A1 * A2 * A3 .. Am,最少要多少次整数乘法? 输入 第一行一个整数n(n <= 100),表示一共有n-1个矩阵.第二行n个整数B1, B2, B3... Bn(Bi <= 100),第i个数Bi表示第i个矩阵的行数和第i-1个矩阵的列数.等价地,可以认为第j个矩阵Aj(1 <= j <= n - 1)的行数为Bj,列数为Bj+1. 输出 一个整数,表示最少所需的乘法次数 采用动…

题目连接:http://poj.org/problem?id=1651 题意:给出一组N个数,每次从中抽出一个数(第一和最后一个不能抽),该次的得分即为抽出的数与相邻两个数的乘积.直到只剩下首尾两个数为止.问最小得分? 分析:区间dp,记忆化搜索,dp[l][r]表示去掉l~r中所有数(不包括l.r)后得到的最小值,那么当前区间最小值为dp[l][r]=min(dp[l][r],dp[l][i]+dp[i][r]+a[l]*a[r]*a[i]). #include <cstdio> #incl…

Multiplication Puzzle Time Limit: 1000MS Memory Limit: 65536K  Total Submissions: 10034 Accepted: 6200 Description The multiplication puzzle is played with a row of cards, each containing a single positive integer. During the move player takes one ca…

比较特别的区间dp.小的区间转移大的区间时,也要枚举断点.不过和普通的区间dp比,断点有特殊意义.表示断点是区间最后取走的点.而且一个区间表示两端都不取走时中间取走的最小花费. #include <iostream> #include <cstdio> #include <cmath> #include <algorithm> #include <vector> #include <iomanip> #include <cstr…

LINK 每次删除一个数,代价是左右两边相邻的数的当前数的积 第一个和最后一个数不能删除 问最后只剩下第一个数的最后一个数的最小代价 思路 很简单的DP 正着考虑没有办法确定两边的数 那么就把每个区间内最后一次删除的数枚举出来,就可以确定左右两边的点是什么了 感觉挺对的 \(dp_{i,j}\)表示还有区间\([i,j]\)全部删完的最小代价,枚举一下最后删除的数然后直接统计贡献就可以了 #include<iostream> #include<cstdio> #include<…

http://blog.csdn.net/libin56842/article/details/9747021 http://www.cnblogs.com/devil-91/archive/2012/06/26/2562976.html #include <iostream> #include <string> #include <cstring> #include <cstdlib> #include <cstdio> #include &l…