After leaving Yemen, Bahosain now works as a salesman in Jordan. He spends most of his time travelling between different cities. He decided to buy a new car to help him in his job, but he has to decide about the capacity of the fuel tank. The new car…

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…

题目链接:Travelling Salesman and Special Numbers 题意: 给出一个二进制数n,每次操作可以将这个数变为其二进制数位上所有1的和(3->2 ; 7->3),现在给出了一个数k,问不大于n的数中有几个数经过k次操作可以变成1. 题解: 因为所给的n很大,但是可以发现只要经过一次操作,n都会变成1000以内的数,所以可以把1000以内的数的答案都存下来.每次在这里面找等于k-1的数,然后数位DP求个数. #include<bits/stdc++.h>…

题目链接:Travelling Salesman and Special Numbers 题意: 给了一个n×m的图,图里面有'N','I','M','A'四种字符.问图中能构成NIMA这种序列最大个数(连续的,比如说NIMANIMA = 2)为多少,如果有环的话那么最大长度就是无穷. 题解: 哇,做了这题深深得感觉自己的dfs真的好弱.这题就是从N开始深搜,在深搜的过程中记录值.返回这个值加1. //#pragma comment(linker, "/stack:200000000"…

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…

Travelling Salesman Problem Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 898    Accepted Submission(s): 327 Special Judge Problem Description Teacher Mai is in a maze with n rows and m colum…

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…

Discription The Travelling Salesman spends a lot of time travelling so he tends to get bored. To pass time, he likes to perform operations on numbers. One such operation is to take a positive integer x and reduce it to the number of bits set to 1 in…

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…

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…