题意: 有一个n * m的数字矩阵,每个格子放着一个非负整数,从左上角走到右下角,每个格子最多走一次,问所经过的格子的最大权值之和是多少,并且输出一个路径. 分析: 如果n和m有一个是偶数的话,那么只要按照蛇形的走法一直走下去即可. 比如n为奇数的时候就这样,左右左右地蛇形走. 同样的,如果m为奇数的时候,也可以上下上下地蛇形走. 但如果n和m都为偶数的时候,就会无法走完全部的格子,最终到达右下角. 但是可以少走一个格子,而且这个格子必须是那种行标加列标为奇数的格子才行(行和列从1开始),所以我…

Travelling Salesman Problem Problem's Link: http://acm.hdu.edu.cn/showproblem.php?pid=5402 Mean: 现有一个n*m的迷宫,每一个格子都有一个非负整数,从迷宫的左上角(1,1)到迷宫的右下角(n,m),并且使得他走过的路径的整数之和最大,问最大和为多少以及他走的路径. analyse: 首先,因为每个格子都是非负整数,而且规定每个格子只能走一次,所以为了使和尽可能大,必定是走的格子数越多越好.这样我们就需…

大致题意:n*m的非负数矩阵,从(1,1) 仅仅能向四面走,一直走到(n,m)为终点.路径的权就是数的和.输出一条权值最大的路径方案 思路:因为这是非负数,要是有负数就是神题了,要是n,m中有一个是奇数.显然能够遍历.要是有一个偶数.能够绘图发现,把图染成二分图后,(1,1)为黑色,总能有一种构造方式能够仅仅绕过不论什么一个白色的点.然后再遍历其它点.而绕过黑色的点必定还要绕过两个白色点才干遍历所有点,这是绘图发现的.所以找一个权值最小的白色点绕过就能够了, 题解给出了证明: ,1)1,而棋盘中…

Travelling Salesman Problem Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 568    Accepted Submission(s): 200 Special Judge Problem Description Teacher Mai is in a maze with n rows and m colum…

行数或列数为奇数就能够所有走完. 行数和列数都是偶数,能够选择空出一个(x+y)为奇数的点. 假设要空出一个(x+y)为偶数的点,则必须空出其它(x+y)为奇数的点 Travelling Salesman Problem Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 747    Accepted Submission(s): 272…

Travelling Salesman Problem PAT-1150 #include<iostream> #include<cstring> #include<string> #include<algorithm> #include<cstdio> #include<sstream> #include<set> #include<map> #include<cmath> #include<…

The "travelling salesman problem" asks the following question: "Given a list of cities and the distances between each pair of cities, what is the shortest possible route that visits each city and returns to the origin city?" It is an N…

https://pintia.cn/problem-sets/994805342720868352/problems/1038430013544464384 The "travelling salesman problem" asks the following question: "Given a list of cities and the distances between each pair of cities, what is the shortest possib…

The "travelling salesman problem" asks the following question: "Given a list of cities and the distances between each pair of cities, what is the shortest possible route that visits each city and returns to the origin city?" It is an N…

Source: PAT A1150 Travelling Salesman Problem (25 分) Description: The "travelling salesman problem" asks the following question: "Given a list of cities and the distances between each pair of cities, what is the shortest possible route that…