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 and the numb…

传送门:http://poj.org/problem?id=1651 Multiplication Puzzle Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13109   Accepted: 8034 Description The multiplication puzzle is played with a row of cards, each containing a single positive intege…

题目链接:http://poj.org/problem?id=1651 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 prod…

题目链接:http://poj.org/problem?id=1651 思路:除了头尾两个数不能取之外,要求把所有的数取完,每取一个数都要花费这个数与相邻两个数乘积的代价,需要这个代价是最小的 用dp[i][j]表示区间[i,j]的最小代价,那么就有dp[i][j]=min(dp[i][k]+dp[k][j]+a[i]*a[k]*a[j]) i+1<=k<=j-1 #include<cstdio> #include<iostream> #include<algor…

题目链接:http://poj.org/problem? id=1651 题意:初使ans=0,每次消去一个值,位置在pos(pos!=1 && pos !=n) 同一时候ans+=a[pos-1]*a[pos]*a[pos+1].一直消元素直到最后剩余2个,求方案最小的ans是多少? 代码: #include <stdio.h> #include <ctime> #include <math.h> #include <limits.h> #…

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…

Multiplication Puzzle Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10010   Accepted: 6188 Description The multiplication puzzle is played with a row of cards, each containing a single positive integer. During the move player takes one…

题目链接:id=1651">点击打开链 题意: 给定一个数组,每次能够选择内部的一个数 i 消除,获得的价值就是 a[i-1] * a[i] * a[i+1] 问最小价值 思路: dp[l,r] = min( dp[l, i] + dp[i, r] + a[l] * a[i] * a[r]); #include <cstdio> #include <iostream> #include <algorithm> #include <queue>…

题意: 给出一个序列,共n个正整数,要求将区间[2,n-1]全部删去,只剩下a[1]和a[n],也就是一共需要删除n-2个数字,但是每次只能删除一个数字,且会获得该数字与其旁边两个数字的积的分数,问最少可以获得多少分数? 思路: 类似于矩阵连乘的问题,用区间DP来做. 假设已知区间[i,k-1]和[k+1,j]各自完成删除所获得的最少分数,那么a[k]是区间a[i,j]内唯一剩下的一个数,那么删除该数字就会获得a[k]*a[i-1]*a[i+1]的分数了.在枚举k的时候要保证[i,j]的任一子区…

E - Multiplication Puzzle  POJ - 1651 这个题目没有特别简单,但是也没有我想象之中的那么难,这个题目时区间dp,因为我们是要对区间进行考虑的. 但是呢,这个也和动态规划的基本原理息息相关. 动态规划 我们一般需要去找这个的子问题,这个题目因为左端点和右端点是不能动 的,所以最后一个取走的就是a[l]*a[k]*a[r] 所以说这个题目的子问题就是每次去找一个区间,然后记录区间的左右端点,然后再找这个区间乘法最小和. dp[i][k]=min(dp[i][k],…

dp题: 1.写状态转移方程; 2.考虑初始化边界,有意义的赋定值.还没计算的赋边界值: 3.怎么写代码自底向上计算最优值 今天做了几个基础dp,所有是dp方程写对可是初始化以及计算写错 先是poj 1651 事实上就是个赤裸裸的矩阵连乘.dp方程非常easy写出 dp[i][j]=min(dp[i][k]+dp[k+1][j]+r[i]*c[k]*c[j],dp[i][j]); 先贴两个个二逼的代码,mark下自己多么的二逼: 二逼一:在计算的时候使用了还没有算出来的值,模拟下就知道第一重循环…

http://poj.org/problem?id=1651 题意:给出n个数字,每取中间一个数,就会使得权值加上中间这个数和两边的乘积,求取剩两个数最少的权值是多少. 思路:区间dp. 一开始想了挺久还是写不出方程,做了点别的事回来再想就突然觉得很简单了. 一开始使得长度为1和2的区间dp[i][j]为0. 然后dp[i][j] = min(dp[i][k] + dp[k][j] + w[k] * w[i] * w[j]). 枚举的k为中间拿掉的数,然后和左右的区间和相加就是最后答案了. #i…

Multiplication Puzzle Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7118   Accepted: 4385 Description The multiplication puzzle is played with a row of cards, each containing a single positive integer. During the move player takes one…

卡片游戏 题目大意:给你一排卡片,你可以从从中抽一些卡片(但是不能抽最左和最右的卡片),每张卡片上有一个数字,当你从中抽出一张卡片后,你将得卡片的数字和卡片左右两张卡片的数字的乘积的分数,问当剩下最左和最右两张卡片的时候,你可以得到的最小的分数? 这一题一看好像挺复杂的,但是如果我们换一个思维方式,这一题就会变得很熟悉 首先这一题是显而易见的DP(找最小值,不能取极限方式) 首先他要我们抽卡片是吧,一开始我们可能会想到先抽一张看看,然后在扫一遍其他卡片看能否抽,但是这样带来的直接后果就是,我们很…

此题可以转化为最优矩阵链乘的形式,d(i, j)表示区间[i, j]所能得到的最小权值. 枚举最后一个拿走的数a[k],状态转移方程为d(i, j) = min{ d(i, k) + d(k, j) + a[i] * a[k] * a[j] } #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; + ; const…

解法 区间dp例题,长度从2开始到n结束起点从1到n,中间枚举的时候是看着左端点右端点与中点的乘积 代码 #include <iostream> #include <cstring> using namespace std; int dp[666][666],num[666]; int main() { ios::sync_with_stdio(0); cin.tie(0); cout.tie(0); int n; cin>>n; memset(dp,0x7f,size…

题目链接: http://poj.org/problem?id=1651 题目大意:加分取牌.如果一张牌左右有牌则可以取出,分数为左牌*中牌*右牌.这样最后肯定还剩2张牌.求一个取牌顺序,使得加分最少. 解题思路: 矩阵链乘的变种题. 假设有10.20.30.40.50五张牌. 如果我想要最后取30,则应该先取20.40,这样就还剩10.30.50三张牌了. 不难发现取20是dp[i][k]部分,取40是dp[k][j]部分,最后剩下的就是i.k.j三张牌. DP边界:无 因为是算加分,区间间隔…

  http://poj.org/problem?id=1651Multiplication Puzzle   Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6188   Accepted: 3777 Description The multiplication puzzle is played with a row of cards, each containing a single positive integer.…

Description Katu Puzzle ≤ c ≤ ). One Katu ≤ Xi ≤ ) such that for each edge e(a, b) labeled by op and c, the following formula holds: Xa op Xb = c The calculating rules are: AND 0 1 0 0 0 1 0 1 OR 0 1 0 0 1 1 1 1 XOR 0 1 0 0 1 1 1 0 Given a Katu Puzzl…

Katu Puzzle Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6553   Accepted: 2401 Description Katu Puzzle is presented as a directed graph G(V, E) with each edge e(a, b) labeled by a boolean operator op (one of AND, OR, XOR) and an integ…

Katu Puzzle Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9987   Accepted: 3741 Description Katu Puzzle is presented as a directed graph G(V, E) with each edge e(a, b) labeled by a boolean operator op (one of AND, OR, XOR) and an integ…

题目链接:http://poj.org/problem?id=3678 代码: #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<vector> using namespace std; ; struct Two_Sat { int n; vector<]; ]; ],cnt; void init(int n) { this-…

Description Katu Puzzle is presented as a directed graph G(V, E) with each edge e(a, b) labeled by a boolean operator op (one of AND, OR, XOR) and an integer c (0 ≤ c ≤ 1). One Katu is solvable if one can find each vertex Vi a value Xi (0 ≤ Xi ≤ 1) s…

                                                                     Katu Puzzle Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11429   Accepted: 4233 Description Katu Puzzle is presented as a directed graph G(V, E) with each edge e(a, …

http://poj.org/problem?id=3678 给m条连接两个点的边,每条边有一个权值0或1,有一个运算方式and.or或xor,要求和这条边相连的两个点经过边上的运算后的结果是边的权值.问存不存在使所有边都符合条件的给点赋值的方法. 2-SAT的各种连法都有了. #include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #include<cmath…

Description A diamond puzzle is played on a tessellated hexagon like the one shown in Figure 1 below. And in this problem the faces produced by the tessellation are identified as they are numbered in the same figure. If two faces share a side, they a…

Description Katu Puzzle is presented as a directed graph G(V, E) with each edge e(a, b) labeled by a boolean operator op (one of AND, OR, XOR) and an integer c (0 ≤ c ≤ 1). One Katu is solvable if one can find each vertex Vi a value Xi (0 ≤ Xi ≤ 1) s…

~~~~ 题目链接:http://poj.org/problem? id=3221 显然是BFS找最优解.但是终止条件不好写.看到有一仅仅队交上去一直TLE. 比赛完了看题解原来是以目标状态为起点,BFS给每一个状态打表.用一个map映射存起来. ~~~~ #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<string> #incl…

Description 给出一个关系,包括 And,Xor,Or 问是否存在解. Sol 经典的2-SAT问题. 把每个值看成两个点,一个点代表选 \(0\) ,另一个代表选 \(1\) . 首先来看 Xor : 如果两个值异或起来为 \(1\) :那么连边 \((i_0,j_1),(i_1,j_0),(j_0,i_1),(j_1,i_0)\) . 否则 连边 \((i_0,j_0),(i_1,j_1),(j_0,i_0),(j_1,i_1)\) . 然后是 And. 如果两个值 And 起来为…

题目链接 题意:给出a, b, c 和操作类型 (与或异或),问是否满足所有的式子 主要是建图: 对于 and , c == 1: 说明 a 和 b都是1,那么 0 就不能取, a' -> a , b' - > b ,因为 a 和 a'是对立事件,对于 a' - >a说明,a'如果成立,那么a也一定存在,显然这是不可能的所以a'不会 成立的. c == 0 说明 a 和 b不全为1, a' -> b , b' -> a 对于 or,  c == 1 :说明 a 和 b 不全为…