1个回答

shiter 回复u014230039: 那你得用qt什么的做做界面把，你说的是控制台版本的程序
5 年多之前 回复

5 年多之前 回复

5 年多之前 回复

c语言迷宫问题怎么解？？

c语言，迷宫问题，救救孩子

Problem Description Alice would like to visit Bob. However, they live in a hilly landscape, and Alice doesn’t like to walk in hills. She has a map of the area, showing the height curves. You have to calculate the total altitude climbed, and the total altitude descended, for the route which minimizes these numbers. It does not matter how far she has to walk to achieve this. Since you don’t know what the landscape looks like in between the height curves, you cannot know exactly how much climb and descent she will actually get in practice, but you should calculate the minimum possible under optimal conditions based on what you can deduce from the map. The map is represented as an xy grid. Alice lives in (0, 0), and Bob lives in (100 000, 0). The height curves are represented as polygons, where a polygon cannot intersect itself or another polygon. Furthermore, neither Alice nor Bob lives exactly on a height curve. Second test case from sample input (compressed). Input On the first line one positive number: the number of testcases, at most 100. After that per testcase: One line with 0 ≤ N ≤ 2 500, the number of height curves. One line for each height curve, with 1 ≤ Hi ≤ 1 000 being the height of the curve, 3 ≤ Pi ≤ 2 000 the number of vertices in the polygon, and the vertices x1, y1, …, xPi, yPi having integral values &#8722;300 000 ≤ xi, yi ≤ 300 000. There will be no more than 200 000 polygon vertices in total in all test cases. Output Per testcase: One line with two numbers: the total altitude climbed and the total altitude descended. Sample Input 2 2 20 3 10 10 0 -10 -10 10 25 3 20 20 0 -20 -20 20 3 100 4 -1 1 1 1 1 -1 -1 -1 300 8 -2 2 2 2 2 -2 5 -2 5 1 6 1 6 -3 -2 -3 50 8 3 3 100001 3 100001 -1 7 -1 7 2 4 2 4 -1 3 -1 Sample Output 5 0 200 250 Source

Problem Description This summer, ELT and his classmates went to Beijing for a training of coding. ELT have never been to Beijing before, so at the weekend, he together with some friends went to the National Museum, it's free for students! The National Museum consists of many parts. One part of it is an exhibition of Ancient China From Xia Dynasty to Qing Dynasty, it needs a big room to show all the things. What's more, there exist many walls to hang pictures. The boundary of this room is walls except the entrance and exit. With walls, an entrance and an exit, this room can be regarded as a maze. To make it simple, this room is a R*C grid, wall is constructed at some edges of grid. The entrance is always at the first row, and the exit is always at the last row, just like the picture below. ELT can't remember his direction in maze, but he is a clever boy. He knew an algorithm called "Always Turn Right", it's procedure is as follows: at any grid of this room, if we can turn right(no wall at right side), then we must turn right; if we can't turn right but can go straight forward, then we must go forward; if we can't go forward but can turn left, then we must turn left; if we can't even turn left, we just turn backward. In the picture above, if we use this algorithm, we'll visit these grids in order: Entrance --> (0, 1) --> (0, 0) --> (0, 1) --> (0, 2) --> (1, 2) --> (1, 1) --> (1, 0) --> (2, 0) --> (1, 0) --> (1, 1) --> (2, 1) --> (2, 2) --> Exit. Very easy, doesn't it? ELT uses "Always Turn Right" algorithm to visit this room from entrance to exit, and then from exit to entrance. He wants to know whether he walked all grids in the room. Now ELT is dizzy because the maze is too big, can you help him? Input First line is an integer T, means T test cases. In each test case, the first line has four numbers: R, C, Ent_Column, Exit_Column. Ent_Column is the column number of entrance; Exit_Column is the column number of exit. Then following 2*R-1 lines, 2*i line have C-1 numbers, the j-th number shows whether there is a wall between grid(i, j) and grid(i, j+1), 2*i+1 line have C numbers, the j-th number shows whether there is a wall between grid(i, j) and grid(i+1, j). Number 1 represents a wall, 0 represents no wall. We guarantee that there exists a path from entrance to exit. 2 <= R, C <= 500 0 <= Ent_Column, Exit_Column < C Output If ELT can walk all grids in the room, print one line "YES", otherwise, print one line "NO". Sample Input 1 3 4 1 2 0 0 0 1 1 0 1 0 0 0 0 0 0 0 1 0 0 Sample Output YES

Problem Description Unlike single maze, double maze requires a common sequence of commands to solve both mazes. See the figure below for a quick understanding. A maze is made up of 6*6 cells. A cell can be either a hole or a square. Moreover, a cell may be surrounded by barriers. There is ONLY one start cell (with a ball) and ONLY one end cell (with a star) in a single maze.These two cells are both squares. It is possible that the start cell and the end cell are the same one. The goal of a single maze is to move the ball from the start cell to the end cell. There are four commands in total,'L', 'D', 'R' and 'U' corresponding to moving the ball left, down, right and up one cell, respectively. The barriers may make the commands take no effect, i.e., the ball does NOT move if there is a barrier on the way. When the ball gets to a hole or outside of the maze, it fails. A double maze is made up of two single mazes. The commands control two balls simultaneously, and the movements of two balls are according to the rules described above independently. Both balls will continue to move simultaneously if at least one of the balls has not got to the end cell. So, a ball may move out of the end cell since the other ball has not been to the target. A double maze passes when both balls get to their end cells, or fails if either of the two mazes fails. The goal of double maze is to get the shortest sequence of commands to pass. If there are multiple solutions, get the lexical minimum one. To simplify the input, a cell is encoded to an integer as follows. The lowest 4 bits signal the existence of the barriers around a cell. The fifth bit indicates whether a cell is a hole or not. The sixth and seventh bits are set for the start cell and end cell. Details are listed in the following table with bits counted from lowest bit. For a barrier, both of the two adjacent cells will have the corresponding barrier bit set. Note that the first two mazes in the sample input is the encoding of two mazes in the figure above, make sure you understand the encoding right. Input The first line of input gives the total number of mazes, T (1 < T ≤ 20). Then follow T mazes. Each maze is a 6*6 matrix, representing the encoding of the original maze. There is a blank line between mazes. Output For every two consecutive mazes, you should treat them as a double maze and output the answer. So there are actually T - 1 answers. For each double maze, output the shortest sequence of commands to pass. If there are multiple solutions, output the lexicographically minimum one. If there is no way to pass, output -1 instead. Sample Input 3 16 0 18 16 18 24 20 19 24 16 28 1 18 28 17 0 22 17 25 20 17 18 88 20 2 16 48 28 17 16 24 16 16 20 23 1 16 0 18 16 18 24 20 19 24 20 29 1 18 28 17 16 22 17 8 20 1 18 24 20 19 80 48 24 16 0 24 16 16 16 22 19 18 16 18 16 18 80 24 18 24 16 24 18 18 24 0 0 18 24 24 18 0 0 24 18 18 24 18 16 18 24 56 18 24 18 24 18 Sample Output RRLULLLRRDLU RURDRLLLURDULURRRRRDDU

Problem Description 度度熊是一只喜欢探险的熊，一次偶然落进了一个m*n矩阵的迷宫，该迷宫只能从矩阵左上角第一个方格开始走，只有走到右上角的第一个格子才算走出迷宫，每一次只能走一格，且只能向上向下向右走以前没有走过的格子，每一个格子中都有一些金币（或正或负，有可能遇到强盗拦路抢劫，度度熊身上金币可以为负，需要给强盗写欠条），度度熊刚开始时身上金币数为0，问度度熊走出迷宫时候身上最多有多少金币？ Input 输入的第一行是一个整数T（T < 200），表示共有T组数据。 每组数据的第一行输入两个正整数m，n（m<=100，n<=100）。接下来的m行，每行n个整数，分别代表相应格子中能得到金币的数量，每个整数都大于等于-100且小于等于100。 Output 对于每组数据，首先需要输出单独一行”Case #?:”，其中问号处应填入当前的数据组数，组数从1开始计算。 每组测试数据输出一行，输出一个整数，代表根据最优的打法，你走到右上角时可以获得的最大金币数目。 Sample Input 2 3 4 1 -1 1 0 2 -2 4 2 3 5 1 -90 2 2 1 1 1 1 Sample Output Case #1: 18 Case #2: 4

Problem Description You are in a rectangular maze and you would like to leave the maze as quickly as possible. The maze is a rectangular grid of square locations. Some locations are blocked. Some other locations are exits. If you arrive at an exit location, you can immediately leave the maze. You may walk one step at a time, onto one of the locations adjacent to your current location. Two locations are adjacent if they share a side. That is, you can only move one step North, South, East or West. Of course, you cannot step off the maze, and you cannot step onto a blocked location. In addition, at any step, you may choose to use your teleport device. This device will send you to a random non-blocked location inside the maze with uniform probability (including, possibly, the one where you currently are standing!). If the device happens to send you onto a spot that is also an exit, then you leave the maze immediately. Hooray! The only way to leave the maze is by moving onto an exit (either by teleport or walking), you may not walk off the boundary of the maze. Write a program to calculate the expected number of steps you need in order to leave the maze. Assume that you would choose your actions (movements and using teleport device) optimally in order to minimize the expected number of steps to leave the maze. Using the teleport device is considered one step. Input There will be multiple test cases. Each test case starts with a line containing two positive integers R and C ( R<=200, C<=200 ). Then, the next R lines each contain C characters, representing the locations of the maze. The characters will be one of the following: E : represents an exit, there will be at least one `E' in every maze. Y : represents your initial location, there will be exactly one `Y' in every maze. X : represents a blocked location. . : represents an empty space. You may move/teleport onto any location that is marked `E', `Y' or `.'. The end of input is marked by a line with two space-separated 0's. Output For each test case, print one line containing the minimum expected number of steps required to leave the maze, given that you make your choices optimally to minimize this value. Print your result to 3 decimal places. Do not print any blank lines between outputs. Sample Input 2 1 E Y 2 2 E. .Y 3 3 EX. XX. ..Y 3 3 EXY .X. ... 0 0

Problem Description 度度熊是一只喜欢探险的熊，一次偶然落进了一个m*n矩阵的迷宫，该迷宫只能从矩阵左上角第一个方格开始走，只有走到右上角的第一个格子才算走出迷宫，每一次只能走一格，且只能向上向下向右走以前没有走过的格子，每一个格子中都有一些金币（或正或负，有可能遇到强盗拦路抢劫，度度熊身上金币可以为负，需要给强盗写欠条），度度熊刚开始时身上金币数为0，问度度熊走出迷宫时候身上最多有多少金币？ Input 输入的第一行是一个整数T（T < 200），表示共有T组数据。 每组数据的第一行输入两个正整数m，n（m<=100，n<=100）。接下来的m行，每行n个整数，分别代表相应格子中能得到金币的数量，每个整数都大于等于-100且小于等于100。 Output 对于每组数据，首先需要输出单独一行”Case #?:”，其中问号处应填入当前的数据组数，组数从1开始计算。 每组测试数据输出一行，输出一个整数，代表根据最优的打法，你走到右上角时可以获得的最大金币数目。 Sample Input 2 3 4 1 -1 1 0 2 -2 4 2 3 5 1 -90 2 2 1 1 1 1 Sample Output Case #1: 18 Case #2: 4
《MySQL 性能优化》之理解 MySQL 体系结构

python爬取百部电影数据，我分析出了一个残酷的真相
2019年就这么匆匆过去了，就在前几天国家电影局发布了2019年中国电影市场数据，数据显示去年总票房为642.66亿元，同比增长5.4%；国产电影总票房411.75亿元，同比增长8.65%，市场占比 64.07%；城市院线观影人次17.27亿，同比增长0.64%。 看上去似乎是一片大好对不对？不过作为一名严谨求实的数据分析师，我从官方数据中看出了一点端倪：国产票房增幅都已经高达8.65%了，为什...

Windows可谓是大多数人的生产力工具，集娱乐办公于一体，虽然在程序员这个群体中都说苹果是信仰，但是大部分不都是从Windows过来的，而且现在依然有很多的程序员用Windows。 所以，今天我就把我私藏的Windows必装的软件分享给大家，如果有一个你没有用过甚至没有听过，那你就赚了????，这可都是提升你幸福感的高效率生产力工具哦！ 走起！???? NO、1 ScreenToGif 屏幕，摄像头和白板...

C语言荣获2019年度最佳编程语言

Idea 中最常用的10款插件（提高开发效率），一定要学会使用！

AI 没让人类失业，搞 AI 的人先失业了

2020年，冯唐49岁：我给20、30岁IT职场年轻人的建议

B站是个宝，谁用谁知道???? 作为一名大学生，你必须掌握的一项能力就是自学能力，很多看起来很牛X的人，你可以了解下，人家私底下一定是花大量的时间自学的，你可能会说，我也想学习啊，可是嘞，该学习啥嘞，不怕告诉你，互联网时代，最不缺的就是学习资源，最宝贵的是啥？ 你可能会说是时间，不，不是时间，而是你的注意力，懂了吧！ 那么，你说学习资源多，我咋不知道，那今天我就告诉你一个你必须知道的学习的地方，人称...

【蘑菇街技术部年会】程序员与女神共舞，鼻血再次没止住。（文末内推）

Java校招入职华为，半年后我跑路了

Python全栈 Linux基础之3.Linux常用命令
Linux对文件（包括目录）有很多常用命令，可以加快开发效率：ls是列出当前目录下的文件列表，选项有-a、-l、-h，还可以使用通配符；c功能是跳转目录，可以使用相对路径和绝对路径；mkdir命令创建一个新的目录，有-p选项，rm删除文件或目录，有-f、-r选项；cp用于复制文件，有-i、-r选项，tree命令可以将目录结构显示出来（树状显示），有-d选项，mv用来移动文件/目录，有-i选项；cat查看文件内容，more分屏显示文件内容，grep搜索内容；>、>>将执行结果重定向到一个文件；|用于管道输出。
​两年前不知如何编写代码的我，现在是一名人工智能工程师

loonggg读完需要5分钟速读仅需 2 分钟大家好，我是你们的校长。我知道大家在家里都憋坏了，大家可能相对于封闭在家里“坐月子”，更希望能够早日上班。今天我带着大家换个思路来聊一个问题...

Spring框架|JdbcTemplate介绍