Travelling Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5762    Accepted Submission(s): 1857 Problem Description After coding so many days,Mr Acmer wants to have a good rest.So travelling is…

http://acm.hdu.edu.cn/showproblem.php?pid=3001 从任何一个点出发,去到达所有的点,但每个点只能到达2次,使用的经费最小.三进制 Travelling Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3541    Accepted Submission(s): 1106 Problem Des…

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7128    Accepted Submission(s): 2297 Problem Description After coding so many days,Mr Acmer wants to have a good rest.So travelling is the best ch…

题目链接:http://acm.split.hdu.edu.cn/showproblem.php?pid=3001 Travelling Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8286    Accepted Submission(s): 2703 Problem Description After coding so many…

题意: 走n个城市, m条路, 起点任意, 每个城市走不超过两次, 求最小花费, 不能走输出-1. $1\le n\le 10$ 分析: 每个城市的拜访次数为0 1 2, 所以三进制状压, 先预处理10位(n最大为10)的三进制数 ], vis[][]; void init() { num[]=; ; i<=; i++) num[i]=num[i-]*; memset(vis, -, sizeof(vis)); ; i<=num[]; i++) { int x=i; ; j<=; j++…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3001 题目: 题意:n个城市,m条边,每条边都有一个权值,问你经过所有的城市且每条边通过次数不超过两次的最短距离. 思路:状压dp+三进制,dp[i][j]表示在状态i下以j为目标城市的最短距离,转移方程为nw = i + fi[k];dp[nw][k] = min(dp[nw][k], dp[i][j] + mp[j][k]). 代码实现如下: #include <set> #include &…

哭瞎啊,每一个城市能够经过至多两次,但没有要求必须经过两次.想用 两个状压来乱搞搞.结果自觉得会T.结果 WA了,搞了一下午.没想到用三进制啊.智商捉急,參考了 http://blog.csdn.net/lenleaves/article/details/7980955  这个博客 每一个城市能够经过1或2次.所以三进制能够代表全部状态了.接下来处理方式类似于二进制的,仅仅是没有了位运算一些推断跟预处理有点繁琐 方程dp[s][i] = min(dp[s][i] ,  dp[s - (s除去j剩…

题目没有起点限制,且每个节点至少访问1次,最多访问2次,所以用三进制数表示节点的状态(选取情况). 因为三进制数的每一位是0或1或2,所以预处理z状态S的第j位的数是有必要的. 边界条件:dp[tri[i]][i]=0,表示只访问了i节点时,从i出发最小费用是0. 最后的答案就在所有满足条件的状态中统计dp[S][u]的最小值,枚举u(因为无起点限制). 1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm&g…

[hdu3001]Travelling Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7817    Accepted Submission(s): 2553 Problem Description After coding so many days,Mr Acmer wants to have a good rest.So trave…

Travelling Salesman   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…

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…

Travelling Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2554    Accepted Submission(s): 746 Problem DescriptionAfter coding so many days,Mr Acmer wants to have a good rest.So travelling is th…

Travelling Tours Time limit: 1.0 secondMemory limit: 64 MB There are N cities numbered from 1 to N (1 ≤ N ≤ 200) and M two-way roads connect them. There are at most one road between two cities. In summer holiday, members of DSAP Group want to make so…

1616: [Usaco2008 Mar]Cow Travelling游荡的奶牛 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 1006  Solved: 548[Submit][Status][Discuss] Description 奶牛们在被划分成N行M列(2 <= N <= 100; 2 <= M <= 100)的草地上游走,试图找到整块草地中最美味的牧草.Farmer John在某个时刻看见贝茜在位置 (R1, C1),恰好T…

1616: [Usaco2008 Mar]Cow Travelling游荡的奶牛 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 762  Solved: 427[Submit][Status] Description 奶牛们在被划分成N行M列(2 <= N <= 100; 2 <= M <= 100)的草地上游走,试图找到整块草地中最美味的牧草.Farmer John在某个时刻看见贝茜在位置 (R1, C1),恰好T (0 < T…

一道水 dp ...然后我一开始用 BFS ...结果 MLE 了... dp[ i ][ j ][ k ] 由它四个方向上的 k - 1 转移. ----------------------------------------------------------------------------------------------- #include<cstdio> #include<algorithm> #include<cstring> #include<…

Travelling Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3905    Accepted Submission(s): 1234 Problem Description After coding so many days,Mr Acmer wants to have a good rest.So travelling is…

Travelling Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6600    Accepted Submission(s): 2144 Problem Description After coding so many days,Mr Acmer wants to have a good rest.So travelling is…

题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=3001 Travelling Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5295    Accepted Submission(s): 1718 Problem Description After coding so many days,…

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…

Travelling Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5896    Accepted Submission(s): 1908 Problem Description After coding so many days,Mr Acmer wants to have a good rest.So travelling is…

Travelling Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8103    Accepted Submission(s): 2642 Problem Description After coding so many days,Mr Acmer wants to have a good rest.So travelling is…

Problem B. Travelling Camera Problem 题目连接: http://www.codeforces.com/gym/100253 Description Programming competitions become very popular in Berland. Now Berland Broadcasting Corporation (BBC) plans to organize a TV broadcast of the All-Berland Region…

TEXT 6 Travelling with baggage 背着行囊去旅行 Feb 16th 2006 From The Economist print edition (1)FEW modern travel writers excite more hostility and awe than Sir ★Wilfred Thesiger[1], who died in 2003. Despising the “drab uniformity of the modern world”, Sir…

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…

http://acm.hdu.edu.cn/showproblem.php?pid=3001 Travelling Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3864    Accepted Submission(s): 1217 Problem Description After coding so many days,Mr A…

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…