梦回高中,定义的f(i,j)为从(0,0)到(i,j)一共有多少条路可以选择,易知我们要做i+j次选择,其中有i次是选择x轴,剩下的是y轴,所以f(i,j)=C(i+j,i)=C(i+j,j),给你一个范围[r1,r2],[c1,c2],求出所有的f(i,j)之和,我们可以用容斥,设g(r,c)为范围[0,r][0,c]的f之和,那么答案就是g(r2,c2)-g(r2,c1-1)-g(r1-1,c2)+g(r1-1,c1-1),目标就转换为快速求g,由组合数公式,g(r,c)就可以从r*c个f和…

Given a binary tree, return all root-to-leaf paths. For example, given the following binary tree: 1 / \ 2 3 \ 5 All root-to-leaf paths are: ["1->2->5", "1->3"] 这道题给我们一个二叉树,让我们返回所有根到叶节点的路径,跟之前那道Path Sum II 二叉树路径之和之二很类似,比那道稍微简单一…

Follow up for "Unique Paths": Now consider if some obstacles are added to the grids. How many unique paths would there be? An obstacle and empty space is marked as 1 and 0 respectively in the grid. For example, There is one obstacle in the middl…

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in t…

Given a binary tree, return all root-to-leaf paths. For example, given the following binary tree: 1 / \ 2 3 \ 5 All root-to-leaf paths are: ["1->2->5", "1->3"] 思路:用两个stack<TreeNode*> in , s; in : 记录当前的路径 p  , 和vector<…

UVA - 10564 Paths through the Hourglass 题意: 要求从第一层走到最下面一层,只能往左下或右下走 问有多少条路径之和刚好等于S? 如果有的话,输出字典序最小的路径. f[i][j][k]从下往上到第i层第j个和为k的方案数 上下转移不一样,分开处理 没必要判断走出沙漏 打印方案倒着找下去行了,尽量往左走   沙茶的忘注释掉文件WA好多次   #include <iostream> #include <cstdio> #include <a…

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in t…

Follow up for "Unique Paths": Now consider if some obstacles are added to the grids. How many unique paths would there be? An obstacle and empty space is marked as 1 and 0 respectively in the grid. For example, There is one obstacle in the middl…

Redundant Paths Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13717   Accepted: 5824 Description In order to get from one of the F (1 <= F <= 5,000) grazing fields (which are numbered 1..F) to another field, Bessie and the rest of the…

题目描述: 1015. Jill's Tour Paths Constraints Time Limit: 1 secs, Memory Limit: 32 MB Description Every year, Jill takes a bicycle tour between two villages. There are different routes she can take between these villages, but she does have an upper limit…

这题在Unique Paths的基础上增加了一些obstacle的位置,应该说增加的难度不大,但是写的时候对细节的要求多了很多,比如,第一列的初始化会受到之前行的第一列的结果的制约.另外对第一行的初始化,也要分if else赋值.很容易出现初始化不正确的情况. 代码: class Solution { public: int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) { ][]==) ; int…

唯一路径问题II Unique Paths II Follow up for "Unique Paths": Now consider if some obstacles are added to the grids. How many unique paths would there be? An obstacle and empty space is marked as 1 and 0 respectively in the grid. For example, There is…

https://leetcode.com/problems/unique-paths/ 这道题,不利用动态规划基本上规模变大会运行超时,下面自己写得这段代码,直接暴力破解,只能应付小规模的情形,当23*12时就超时了: class Solution { public: // Solution():dp1(m,vector<int>(n,-1)),dp2(m,vector<int>(n,-1)){ // } int uniquePaths(int m, int n) { helper…

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in t…

Paths中的几个重要元素 Points void CGContextMoveToPoint ( CGContextRef c, CGFloat x, CGFloat y ); 指定一个点成为current point Quartz会跟踪current point一般执行完一个相关函数后,current point都会相应的改变. Lines 相关的几个函数 void CGContextAddLineToPoint ( CGContextRef c, CGFloat x, CGFloat y )…

Given a binary tree, return all root-to-leaf paths. For example, given the following binary tree: 1 / \ 2 3 \ 5 All root-to-leaf paths are: ["1->2->5", "1->3"] 本题可以用DFS或者BFS. 解法一: DFS,参考自九章算法 class Solution: # @param {TreeNode…

渲染路径 Rendering Paths http://game.ceeger.com/Manual/RenderingPaths.html 延迟光照渲染路径的细节 Deferred Lighting Rendering Path Details http://game.ceeger.com/Components/RenderTech-DeferredLighting.html 正向渲染路径细节 Forward Rendering Path Details http://game.ceeger.…

git commit -am '*屏蔽设置缓存' htdocs/s.php fatal: Paths with -a does not make sense. 应该用下面的这样. git commit -m '*屏蔽设置缓存' htdocs/s.php…

Follow up for "Unique Paths": Now consider if some obstacles are added to the grids. How many unique paths would there be? An obstacle and empty space is marked as 1 and 0 respectively in the grid. For example, There is one obstacle in the middl…

Unique Paths II Total Accepted: 22828 Total Submissions: 81414My Submissions Follow up for "Unique Paths": Now consider if some obstacles are added to the grids. How many unique paths would there be? An obstacle and empty space is marked as 1 an…

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in t…

Given a binary tree, return all root-to-leaf paths.Example Given the following binary tree: 1 /   \2     3 \  5 All root-to-leaf paths are: [  "1->2->5",  "1->3"] LeetCode上的原题,请参见我之前的博客Binary Tree Paths. 解法一: class Solution {…

NIO.2 JDK7对NIO进行了重大改进,主要包含以下两方面 新增Path接口,Paths工具类,Files工具类. 这些接口和工具类对NIO中的功能进行了高度封装,大大简化了文件系统的IO编程. 基于异步Channel的IO 在NIO基础上改进后的IO被称为NIO.2 , 上面第一个改进包含在java.nio下新增的包java.nio.file包. 第二个改进包含在原有的java.nio.channels下,新增了多个Aysnchronous开头的channel接口和类. 本文暂时只讨论第一…

Redundant Paths Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13712   Accepted: 5821 Description In order to get from one of the F (1 <= F <= 5,000) grazing fields (which are numbered 1..F) to another field, Bessie and the rest of the…

Binary Tree Paths Given a binary tree, return all root-to-leaf paths. Given the following binary tree: 1 / \ 2 3 \ 5 All root-to-leaf paths are: [ "1->2->5", "1->3" ] As mentioned in the problem we want to get all root-to-leaf…

Unique Paths A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked…

Binary Tree Paths Given a binary tree, return all root-to-leaf paths. For example, given the following binary tree: 1 / \ 2 3 \ 5 All root-to-leaf paths are: ["1->2->5", "1->3"] Credits:Special thanks to @jianchao.li.fighter f…

Unique Paths II Total Accepted: 31019 Total Submissions: 110866My Submissions Question Solution  Follow up for "Unique Paths": Now consider if some obstacles are added to the grids. How many unique paths would there be? An obstacle and empty spa…

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in t…

Java7中文件IO发生了很大的变化,专门引入了很多新的类: import java.nio.file.DirectoryStream;import java.nio.file.FileSystem;import java.nio.file.FileSystems;import java.nio.file.Files;import java.nio.file.Path;import java.nio.file.Paths;import java.nio.file.attribute.FileAt…