AnatBird.com COST CUTTING THE ALAN GREENBERG WAY…

题号:a001: 哈囉 背景知识:输出语句,while not eof 题号:a002: 簡易加法 背景知识:输出语句,while not eof,加法运算 题号:a003: 兩光法師占卜術 背景知识:while not eof,分支语句,求余 PS:%表示求余 题号:a004: 文文的求婚 背景知识:while not eof,分支语句,判断闰年 题号:a005: Eva 的回家作業 背景知识:循环,if语句,等比数列,等差数列,输出语句 题号:a006: 一元二次方程式 背景知识:if语句的…

Unit 1 overview of IT IndustryConcept LearningIT Industry OutlookThe term technology commonly refers to society’s application of scientific knowledge to solvepractical problems in industry or commerce. Technological innovation, or the application oft…

  Cutting Sticks  You have to cut a wood stick into pieces. The most affordable company, The Analog Cutting Machinery, Inc. (ACM), charges money according to the length of the stick being cut. Their procedure of work requires that they only make one…

题意:在给出的n个结点处切断木棍,并且在切断木棍时木棍有多长就花费多长的代价,将所有结点切断,并且使代价最小. 思路:设DP[i][j]为,从i,j点切开的木材,完成切割需要的cost,显然对于所有DP[i][i+1]=0,记w[i][j]为从i,j点切开的木材的长度,因此可以枚举切割点,dp[i][j]=min(dp[i][k]+dp[k][j])+w[i][j],k就是枚举的切割点. AC代码: #include<cstdio> #include<cstring> #inclu…

Cutting There are a lot of things which could be cut — trees, paper, “the rope”. In this problem you are going to cut a sequence of integers. There is a sequence of integers, which contains the equal number of even and odd numbers. Given a limited bu…

Tree Cutting Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 4834   Accepted: 2958 Description After Farmer John realized that Bessie had installed a "tree-shaped" network among his N (1 <= N <= 10,000) barns at an incredible…

之前发过一篇二级指针操作单向链表的例子,显示了C语言指针的灵活性,这次再探讨一个指针操作链表的例子,而且是一种完全不同的用法. 这个例子是linux-1.2.13网络协议栈里的,关于链表遍历&数据拷贝的一处实现.源文件是/net/inet/dev.c,你可以从kernel.org官网上下载. 从最早的0.96c版本开始,linux网络部分一直采取TCP/IP协议族实现,这是最为广泛应用的网络协议,整个架构就是经典的OSI七层模型的描述,其中dev.c是属于链路层实现.从功能上看,其位于网络设备驱…

The Benefits of Cutting Salt 减少盐分摄取量的益处 ①Just when you had figured out how to manage fat in your diet,researchers are now warning against another common mealtime pit-fall—salt. ②A study by researchsers at the University of California,San Francisco(UC…

After Farmer John realized that Bessie had installed a "tree-shaped" network among his N (1 <= N <= 10,000) barns at an incredible cost, he sued Bessie to mitigate his losses. Bessie, feeling vindictive, decided to sabotage Farmer John's n…

Description    Cutting Sticks  You have to cut a wood stick into pieces. The most affordable company, The Analog Cutting Machinery, Inc. (ACM), charges money according to the length of the stick being cut. Their procedure of work requires that they o…

怀着很纠结的心情来总结这篇论文,这主要是因为作者提虽然供了源代码,但是我并没有仔细去深究他的code,只是把他的算法加进了自己的项目.希望以后有时间能把MST这一结构自己编程实现!! 论文题目是基于非局部代价聚类(non-local cost aggregation)的立体匹配,从题目上看这篇论文不是局部算法,但是也不是传统意义上的全局算法.这要从基于窗结构局部立体匹配算法说起,如下图: 我们求左右两幅图像在视差d下一点的cost时,我们实际是求得以该点为中心半径为r的Windows内所有点的c…

卧槽....最近刷的cf上有最短路,本来想拿这题复习一下.... 题意就是在输出最短路的情况下,经过每个节点会增加税收,另外要字典序输出,注意a到b和b到a的权值不同 然后就是处理字典序的问题,当松弛时发现相同值的时候,判断两条路径的字典序 代码 #include "stdio.h" const int MAXN=110; const int INF=10000000; bool vis[MAXN]; int pre[MAXN]; int cost[MAXN][MAXN],lowcos…

实际上,代价函数(cost function)和损失函数(loss function 亦称为 error function)是同义的.它们都是事先定义一个假设函数(hypothesis),通过训练集由算法找出一个最优拟合,即通过使的cost function值最小(如通过梯度下降),从而估计出假设函数的未知变量. 例如: 可以看做一个假设函数,而与之对应的loss function如下: 通过使E(w)值最小,来估计出相应的w值,从而确定出假设函数(目标函数),实现最优拟合. 硬要说区别的话,l…

有向图中N*N矩阵 cost:M, 最多可以遍历的结点个数 例如A可以有0->1->2->0->1 代价:2+2+3+2=9<10 输出4 #include <iostream> using namespace std; int main(){ ;; ]={{,,},{,,},{,,}}; ]; ][]; ;i<N;i++) { ;j<N;j++) { dist[i][j]=A[j][i]; } } ;i<N;i++){last_step[i]=…

@tags: caffe 机器学习 在机器学习(暂时限定有监督学习)中,常见的算法大都可以划分为两个部分来理解它 一个是它的Hypothesis function,也就是你用一个函数f,来拟合任意一个输入x,让预测值t(t=f(x))来拟合真实值y 另一个是它的cost function,也就是你用一个函数E,来表示样本总体的误差. 而有时候还会出现loss function,感觉会和cost function混淆. 上quora看了下,有个同名问题,回答的人不多,upvote更少..回答者里面…

Mean Square Error \[cost(t,o)=\frac{1}{n}\sum\limits_{i=1}^n{(o-t)^2}\] Binary Cross-Entropy 用于计算 target 和 output 之间的binary 交叉熵.\[cost(t,o)=-{[t\ln(o) + (1-t)\ln (1-o)]}\] 也可以写作: \[cost(t,o)=\left\{\begin{array}{*{20}{l}}{\ln o, ~~~~~~t=1,}\\ {\ln (1…

Given an integer array, adjust each integers so that the difference of every adjacent integers are not greater than a given number target. If the array before adjustment is A, the array after adjustment is B, you should minimize the sum of |A[i]-B[i]…

https://vjudge.net/problem/11079/origin 题意:有N个商店和M个供应商和K种物品,每个商店每种物品有一个需求数,每个供应商每种物品有一个供应量,供应商到商店之间的运输需要花费,如果供不应求输出-1,否则输出最小花费. 思路:比较明显的最小费用流.想法大概都是源点和供应商连一条容量为供应量,花费为0的边,商店和汇点之间连一条容量为需求量,花费为0的边,供应商和商店之间连一条容量为INF,花费为题意给出的花费的边.建图的话一开始是直接对于每一个商店每一种物品和每…

Description In order to build a ship to travel to Eindhoven, The Netherlands, various sheet metal parts have to be cut from rectangular pieces of sheet metal. Each part is a convex polygon with at most 8 vertices. Each rectangular piece of sheet meta…

貌似···················· 这个算法深的东西还是很不熟悉!继续学习!!!! ++++++++++++++++++++++++++++ ============================ ++++++++++++++++++++++++++++ ------------------------------------------------- ============================ #include<stdio.h> #include<string…

Minimum Transport Cost Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9109    Accepted Submission(s): 2405 Problem Description These are N cities in Spring country. Between each pair of cities…

题目传送门 /* 贪心水题:首先,最少的个数为n最大的一位数字mx,因为需要用1累加得到mx, 接下来mx次循环,若是0,输出0:若是1,输出1,s[j]--: 注意:之前的0的要忽略 */ #include <cstdio> #include <iostream> #include <cstring> #include <string> #include <algorithm> #include <cmath> #include &…

题目传送门 /* 水题:一开始看错题意,以为是任意切割,DFS来做:结果只是在中间切出一段来 判断是否余下的是 "CODEFORCES" :) */ #include <cstdio> #include <iostream> #include <cstring> #include <string> #include <algorithm> #include <cmath> #include <set>…

一开始我把每个店主都拆成k个点,然后建图..然后TLE.. 看题解= =哦,愚钝了,k个商品是独立的,可以分别跑k次最小费用最大流,结果就是k次总和.. #include<cstdio> #include<cstring> #include<queue> #include<algorithm> using namespace std; #define INF (1<<30) #define MAXN 111 #define MAXM 111*11…

题目链接: http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3626 题目大意:树中取点.每过一条边有一定cost,且最后要回到起点.给定预算m,问最大价值. 解题思路: 首先要注意这题要回到起点,由于树的特殊结构(每个结点只有一个父亲)也就是说,要回到开头, 开销是2倍.所以首先m/=2. 然后就是树形背包的求解,这题的cost在边,所以for写法变成如下: for(m....j....0)     for(0....k…

http://accountingexplained.com/financial/inventories/avco-method   Average Cost (AVCO) Method   Average cost method (AVCO) calculates the cost of ending inventory and cost of goods sold for a period on the basis of weighted average cost per unit of i…

Given an integer array, adjust each integers so that the difference of every adjcent integers are not greater than a given number target. If the array before adjustment is A, the array after adjustment is B, you should minimize the sum of |A[i]-B[i]|…

Cutting Game 题意: 有一张被分成 w*h 的格子的长方形纸张,两人轮流沿着格子的边界水平或垂直切割,将纸张分割成两部分.切割了n次之后就得到了n+1张纸,每次都可以选择切得的某一张纸再进行切割.最先切出只有一个格子的纸张(即有 1*1 格子的)的一方获胜.当双方都采取最优策略时,先手必胜还是必败? 题解: 这题是小白书上的,在取得游戏中必胜策略的最后一题,我照着码一遍就叫了,结果居然T了,,最后发现是cin的问题,然后关下同步就可以a了,但好险用了913ms... 还有初始化mem…

题意略: 思路: 这题比较坑的地方是把每种货物单独建图分开算就ok了. #include<stdio.h> #include<queue> #define MAXN 500 #define MAXM 10002*4 #define INF 10000000 using namespace std; //起点编号必须最小,终点编号必须最大 bool vis[MAXN]; //spfa中记录是否在队列里边 ][]; int max_flow; struct point{ int x,y…