题目链接: http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2027 题目: Samball is going to travel in the coming vacation. Now it's time to make a plan. After choosing the destination city, the next step is to determine the travel route. As this…

// 题意 : 一个人要去旅行 给你起点和终点 求最少花费 其中花费为经过路径的总费用减去该路径的中的最大花费段// 直接搜索 稍微加了个剪枝 主要是数据规模小#include <iostream> #include <map> #include <algorithm> #include <queue> #include <math.h> #include <stdio.h> #include <string.h> #in…

给一张有向无环图,边都有花费,从某点到某点走的那条路径上的那一条花费最多的边可以省掉,问从起点到终点的最少花费的多少, 往DP想的话,就可以写出这个状态dp[u][mx],表示到达u点已经省掉的花费为mx的最少花费. 用SPFA更新转移方程..或者理解成队列+我为人人的转移..其实这题这样子也能解有环图. 看了别人博客,发现还有三种解法: 枚举每一条边作为省掉的边,n次SPFA.这方法简洁,可惜想不出= = 跑Dijkstra,根据记录到每一点时的最长边更新,正确性不懂.. Floyd+DP:加…

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…

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…

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,…

[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…

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…

Travelling After coding so many days,Mr Acmer wants to have a good rest.So travelling is the best choice!He has decided to visit n cities(he insists on seeing all the cities!And he does not mind which city being his start station because superman can…

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…

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…

//POJ3377 //DP解法-解有规律的最短路问题 //Time:1157Ms Memory:12440K #include<iostream> #include<cstring> #include<cstdio> using namespace std; #define MAXN 1000005 typedef long long LL; int n; int dp[MAXN][3]; int sr, st, er, ed; int main() { //freo…

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…

FEE Development Essentials JS Basic function call() and apply() func1.bind(thisObj,arg1...argn) Custom object prototype Serialize object via JSON functions Object oriented in javascript Inheritance APIs Nodejs Global object and variable Module Core m…

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<…

一个人的旅行 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 33401    Accepted Submission(s): 11492 Problem Description 虽然草儿是个路痴(就是在杭电待了一年多,居然还会在校园里迷路的人,汗~),但是草儿仍然很喜欢旅行,因为在旅途中 会遇见很多人(白马王子,^0^),很多事,还能丰…

Your are given an array of integers prices, for which the i-th element is the price of a given stock on day i; and a non-negative integer fee representing a transaction fee. You may complete as many transactions as you like, but you need to pay the t…

题目: 在上一篇博客的基础上,这是另一种方法求最短路径的问题. Dijkstra(迪杰斯特拉)算法:找到最短距离已经确定的点,从它出发更新相邻顶点的最短距离.此后不再关心前面已经确定的“最短距离已经确定的点”. Dijkstra算法采用的是一种贪心的策略,声明一个数组dis来保存源点到各个顶点的最短距离和一个保存已经找到了最短路径的顶点的集合:T,初始时,原点 s 的路径权重被赋为 0 (dis[s] = 0).若对于顶点 s 存在能直接到达的边(s,m),则把dis[m]设为w(s, m),同…

题目: 最短路:给定两个顶点,在以这两个点为起点和终点的路径中,边的权值和最小的路径.考虑权值为点之间的距离. 单源最短路问题,Bellman-ford算法 思路:每次循环检查所有边,可优化. 应用于旅游等路径最小问题. 代码: import java.util.Arrays; public class 图的最短路问题_单源 { public static void main(String[] args) { int[] shortestPath = shortestPath(0); Syste…

简介 最近这段时间刚好做了最短路问题的算法报告,因此对dijkstra算法也有了更深的理解,下面和大家分享一下我的学习过程. 前言 呃呃呃,听起来也没那么难,其实,真的没那么难,只要弄清楚思路就很容易了.下面正经的跟大家说下解决问题的过程. 实现过程 我们先用一个d[i]数组表示起点到点i的直接距离,然后从d[i]数组中找最小的值所对应的点,然后看点与点i之间相连的点j, 然后比较d[j]和d[i]+w[i][j](w[i][j]表示的是点i到点j之间的距离)之间的大小,然后把d[j]和d[i]…

Your are given an array of integers prices, for which the i-th element is the price of a given stock on day i; and a non-negative integer fee representing a transaction fee. You may complete as many transactions as you like, but you need to pay the t…

最短路问题之 Floyd 某公司在六个城市 c1c1,c2c2,….,c6c6 中有分公司,从 cici 到 cjcj 的直接航程票价记在下述矩阵的 (ii,jj) 位置上. (∞∞表示无直接航路),请帮助该公司设计一张城市 c1c1 到其它城市间的票价便宜的路线图. 变量解释: n 是公司个数 a 存储航路票价,最后结束循环存储的是最便宜票价 path 存储每对顶点之间最短路径上所经过的定点的序号,也就是”中转站”序号 clear;clc; n = 6; a = [0 50 inf 40 25…

最短路问题(Bellman/Dijkstra/Floyd) 寒假了,继续学习停滞了许久的算法.接着从图论开始看起,之前觉得超级难的最短路问题,经过两天的苦读,终于算是有所收获.把自己的理解记录下来,可以加深印象,并且以后再忘了的时候可以再看.最短路问题在程序竞赛中是经常出现的内容,解决单源最短路经问题的有bellman-ford和dijkstra两种算法,其中,dijikstra算法是对bellman的改进.解决任意两点间的最短路有Floyd-warshall算法. 单源最短路1(bellman…

题目链接: https://cn.vjudge.net/problem/POJ-3037 Bessie and the rest of Farmer John's cows are taking a trip this winter to go skiing. One day Bessie finds herself at the top left corner of an R (1 <= R <= 100) by C (1 <= C <= 100) grid of elevati…