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价值即等价于给每一个点系数$p_{i}=\pm 1$,使得$\forall (x,y)\in E,p_{x}=p_{y}$的最大的$\sum_{i=1}^{n}p_{i}b_{i}$ 如果没有删除(当然可以直接求绝对值),考虑网络流建图:将$b_{i}$分为正负两类,$S$向正的连$2b_{i}$的边,负的向$T$连$-2b_{i}$的边,将图中直接相连的两点连上流量为$\infty$的边,那么若两点不同向,则必须有一个被割掉 考虑删除,可以看作这个这个点不要求正负,同时其也不会帮助连边,因此将…
Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution. For example, given array S = {- -},…
Problem 1603 - Minimum Sum Time Limit: 2000MS   Memory Limit: 65536KB    Total Submit: 563  Accepted: 156  Special Judge: No Description There are n numbers A[1] , A[2] .... A[n], you can select m numbers of it A[B[1]] , A[B[2]] ... A[B[m]]  ( 1 <= B…
Given an array nums of n integers and an integer target, find three integers in nums such that the sum is closest to target. Return the sum of the three integers. You may assume that each input would have exactly one solution. Example: Given array nu…
Problem 1603 - Minimum Sum Time Limit: 2000MS   Memory Limit: 65536KB   Total Submit: 623  Accepted: 178  Special Judge: No Description There are n numbers A[1] , A[2] .... A[n], you can select m numbers of it A[B[1]] , A[B[2]] ... A[B[m]]  ( 1 <= B[…
Array 448.找出数组中所有消失的数 要求:整型数组取值为 1 ≤ a[i] ≤ n,n是数组大小,一些元素重复出现,找出[1,n]中没出现的数,实现时时间复杂度为O(n),并不占额外空间 思路1:(discuss)用数组下标标记未出现的数,如出现4就把a[3]的数变成负数,当查找时判断a的正负就能获取下标 tips:注意数组溢出 public List<Integer> findDisappearedNumbers(int[] nums) { List<Integer> d…
Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution. For example, given array S = {-1 2…
​ 这里我们介绍的是 40+ 个非常有用的 Oracle 查询语句,主要涵盖了日期操作,获取服务器信息,获取执行状态,计算数据库大小等等方面的查询.这些是所有 Oracle 开发者都必备的技能,所以快快收藏吧! 日期/时间 相关查询 获取当前月份的第一天 运行这个命令能快速返回当前月份的第一天.你可以用任何的日期值替换 “SYSDATE”来指定查询的日期. 1 SELECT TRUNC (SYSDATE, 'MONTH') "First day of current month"  2…
上计算机视觉课老师布置的作业实现论文:Color Transfer between Images 基本思路是: 1.给定srcImg和targetImg 2.将RGB空间转为Lab空间 3.根据论文中公式: 计算每一个像素点 4.将resultImg转回到RGB空间显示 效果图:       见代码: #include <opencv2/opencv.hpp> #include <opencv2/core/core.hpp> #include <opencv2/imgproc…