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1. https://www.mrpt.org/Building_and_Installing_Instructions#1_Prerequisites P1. error C2371: “int32_t”: 重定义:不同的基类型  或“int8_t” 解决办法:因为两个.h文件所定义的int32_t和int8_t的类型不同.错误会提示哪两个.h文件冲突,打开pstdint.h文件,找到对应的定义,并修改为另一个.h文件的定义类型. P2. Miscellaneous.h文件 error C27…
Problem UVA12569-Planning mobile robot on Tree (EASY Version) Accept:138  Submit:686 Time Limit: 3000 mSec  Problem Description  Input The first line contains the number of test cases T (T ≤ 340). Each test case begins with four integers n, m, s, t (…
基本思路就是Bfs: 本题的一个关键就是如何判段状态重复. 1.如果将状态用一个int型数组表示,即假设为int state[17],state[0]代表机器人的位置,从1到M从小到大表示障碍物的位置.那么如果直接用STL中的set是会超时的,但如果自己建立一个hash方法,像这样: int getKey(State& s) { long long v = 0; for(int i=0; i<=M; ++i ) { v = v * 10 + s[i]; } return v % hashSi…
题目大意:一张无向连通图,有一个机器人,若干个石头,每次移动只能移向相连的节点,并且一个节点上只能有一样且一个东西(机器人或石头),找出一种使机器人从指定位置到另一个指定位置的最小步数方案,输出移动步骤. 题目分析:以机器人的所在位置和石头所在位置集合标记状态,状态数最多有15*2^15个.广搜之. 代码如下: # include<iostream> # include<cstdio> # include<string> # include<queue> #…
用(x,s)表示一个状态,x表示机器人的位置,s表示其他位置有没有物体.用个fa数组和act数组记录和打印路径,转移的时候判断一下是不是机器人在动. #include<bits/stdc++.h> using namespace std; ; ; +; // 2^15*15 int head[maxn],to[maxe],nxt[maxe]; int ecnt; void addEdge(int u,int v) { to[ecnt] = v; nxt[ecnt] = head[u]; hea…
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Tum RGB-D SLAM Dataset and Benchmark https://vision.in.tum.de/data/datasets/rgbd-dataset Kitti http://www.cvlibs.net/datasets/kitti/index.php CMU Dataset collected using the Navlab 11 equipped with IMU, GPS, Lidars and cameras http://3dvis.ri.cmu.edu…
原文链接 Awesome C++ A curated list of awesome C++ (or C) frameworks, libraries, resources, and shiny things. Inspired by awesome-... stuff. Awesome C++ Standard Libraries Frameworks Artificial Intelligence Asynchronous Event Loop Audio Biology BitTorren…