题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1690 Bus System Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6569    Accepted Submission(s): 1692 Problem Description Because of the huge popula…

Problem Description Because of the huge population of China, public transportation is very important. Bus is an important transportation method in traditional public transportation system. And it's still playing an important role even now. The bus sy…

Bus System Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6082    Accepted Submission(s): 1560 Problem Description Because of the huge population of China, public transportation is very importa…

题目大意:给出若干巴士不同价格的票的乘坐距离范围,现在有N个站点,有M次询问,查询任意两个站点的最小花费 解析:由于是多次查询不同站点的最小花费,所以用弗洛伊德求解 时间复杂度(O^3) 比较基础的弗洛伊德 #include <iostream> #include <cstdio> #include <cstring> using namespace std; #define INF 1000000000000 typedef __int64 LL; ; __int64…

Bus System Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5190    Accepted Submission(s): 1275 [题目链接]http://acm.hdu.edu.cn/showproblem.php?pid=1690 [解题思路]SPFA求最短路径问题,将每个站之间的距离转化为相应的价格从而建立两站点相连的…

Problem Description Because of the huge population of China, public transportation is very important. Bus is an important transportation method in traditional public transportation system. And it’s still playing an important role even now.The bus sys…

Problem Description Because of the huge population of China, public transportation is very important. Bus is an important transportation method in traditional public transportation system. And it’s still playing an important role even now. The bus sy…

Bus Pass Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 667    Accepted Submission(s): 271 Problem Description You travel a lot by bus and the costs of all the seperate tickets are starting to…

Dijkstra最短路算法 --转自啊哈磊[坐在马桶上看算法]算法7:Dijkstra最短路算法 上节我们介绍了神奇的只有五行的Floyd最短路算法,它可以方便的求得任意两点的最短路径,这称为“多源最短路”.本周来来介绍指定一个点(源点)到其余各个顶点的最短路径,也叫做“单源最短路径”.例如求下图中的1号顶点到2.3.4.5.6号顶点的最短路径. 与Floyd-Warshall算法一样这里仍然使用二维数组e来存储顶点之间边的关系,初始值如下. 我们还需要用一个一维数组dis来存储1号顶点到其余各…

dijkstra(最短路)和Prim(最小生成树)下的堆优化 最小堆: down(i)[向下调整]:从第k层的点i开始向下操作,第k层的点与第k+1层的点(如果有)进行值大小的判断,如果父节点的值大于子节点的值,则修改,并继续对第k+1层与第k+2层的点进行判断和修改,否则不修改,且退出.当点向下移动到树的最后一层,没有子节点供判断与修改,停止操作. 树最多有log(n) 层[log(n)=log2n,一般省略数字2],时间复杂度log(n)次. up(i)[向上调整]:同理,时间复杂度log(…

       上周我们介绍了神奇的只有五行的Floyd最短路算法,它可以方便的求得任意两点的最短路径,这称为“多源最短路”.本周来来介绍指定一个点(源点)到其余各个顶点的最短路径,也叫做“单源最短路径”.例如求下图中的1号顶点到2.3.4.5.6号顶点的最短路径.           与Floyd-Warshall算法一样这里仍然使用二维数组e来存储顶点之间边的关系,初始值如下.           我们还需要用一个一维数组dis来存储1号顶点到其余各个顶点的初始路程,如下.          …

Bus System Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8055    Accepted Submission(s): 2121 Problem Description Because of the huge population of China, public transportation is very importa…

hdu 5552 Bus Routes 考虑有环的图不方便,可以考虑无环连通图的数量,然后用连通图的数量减去就好了. 无环连通图的个数就是树的个数,又 prufer 序我们知道是 $ n^{n-2} $ 其中又由于有 $ n-1 $ 个边,每个边可以涂色,所以总共无环的方案数量是 $ m^{n-1} n^{n-2} $ 那么现在就要算连通图的数量了.这个不如不连通图的数量好算. 不连通图的数量怎么算呢,原本想的是容斥,但是貌似不好实现,看了题解发现一种神仙思路.考虑固定一个点,并且让这个点连出一…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2544 Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Problem Description 在每年的校赛里,所有进入决赛的同学都会获得一件很漂亮的t-shirt.但是每当我们的工作人员把上百件的衣服从商店运回到赛场的时候,却是非常累的!所以现在他们想要寻找最短的从商店到赛场的路线,你可…

Problem Description Gabiluso is one of the greatest spies in his country. Now he's trying to complete an "impossible" mission ----- to make it slow for the army of City Colugu to reach the airport. City Colugu has n bus stations and m roads. Eac…

http://acm.hdu.edu.cn/showproblem.php?pid=1548 Online Judge Online Exercise Online Teaching Online Contests Exercise Author F.A.QHand In HandOnline Acmers Forum |DiscussStatistical Charts Problem ArchiveRealtime Judge StatusAuthors Ranklist       C/C…

#include<stdio.h> #include<string.h> #include<queue> #include<algorithm> using namespace std; typedef __int64 ll; #define inf 0x3f3f3f3f //#define inf 0x7f7f7f7f7f7f7f7fLL #define min(x,y) (x<y?x:y) ll l1,l2,l3,l4,c1,c2,c3,c4; i…

题目链接: 传送门 Invitation Cards Time Limit: 5000MS     Memory Limit: 32768 K Description In the age of television, not many people attend theater performances. Antique Comedians of Malidinesia are aware of this fact. They want to propagate theater and, mo…

Telephone Lines Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7214   Accepted: 2638 Description Farmer John wants to set up a telephone line at his farm. Unfortunately, the phone company is uncooperative, so he needs to pay for some of…

hdu 2544  求点1到点n的最短路  无向图 Sample Input2 1 //结点数 边数1 2 3 //u v w3 31 2 52 3 53 1 20 0 Sample Output32 堆优化Dijstra模板 # include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # include <cmath> # include &…

Domino Effect Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9335   Accepted: 2325 Description Did you know that you can use domino bones for other things besides playing Dominoes? Take a number of dominoes and build a row by standing t…

题目链接:[http://acm.hdu.edu.cn/showproblem.php?pid=3339] In Action Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5951    Accepted Submission(s): 1998 Problem Description Since 1945, when the firs…

[题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=6071 [题目大意] 给出四个点1,2,3,4,1和2,2和3,3和4,4和1 之间有路相连, 现在从2点出发,最后回到2点,要求路径大于等于K,问路径长度最短是多少 [题解] 取一条与2相连的权值最小的边w. 若存在一条从起点到终点的长度为k的路径, 那么必然存在一条长度为k+2w的路径,只要一开始在那条边上往返走就好了. 设dij表示从起点到i,路径长度模2w为j时,路径长度的最小值. 用最短…

Bus Fair Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 600    Accepted Submission(s): 293 Problem Description You are now in Foolish Land. Once moving in Foolish Land you found that there is…

Tricks Device 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5294 Description Innocent Wu follows Dumb Zhang into a ancient tomb. Innocent Wu's at the entrance of the tomb while Dumb Zhang's at the end of it. The tomb is made up of many chambers, th…

http://acm.hdu.edu.cn/showproblem.php?pid=5521 Meeting Problem Description   Bessie and her friend Elsie decide to have a meeting. However, after Farmer John decorated hisfences they were separated into different blocks. John's farm are divided into …

题目描述 现在是晚餐时间,而母牛们在外面分散的牧场中. 农民约翰按响了电铃,所以她们开始向谷仓走去. 你的工作是要指出哪只母牛会最先到达谷仓(在给出的测试数据中,总会有且只有一只最快的母牛). 在挤奶的时候(晚餐前),每只母牛都在她自己的牧场上,一些牧场上可能没有母牛. 每个牧场由一条条道路和一个或多个牧场连接(可能包括自己). 有时,两个牧场(可能是字母相同的)之间会有超过一条道路相连. 至少有一个牧场和谷仓之间有道路连接. 因此,所有的母牛最后都能到达谷仓,并且母牛总是走最短的路径. 当然,…

最朴素的做法o(V*V/2+2E)~O(V^2)#include<iostream>using namespace std;#include<vector>#include<algorithm>#include<string>#include<string.h>const int MAX = 2002;int n;int graph[MAX][MAX];int dis[MAX];bool vis[MAX];const int INF = 0X7F…

HDU 3790 最短路径问题 Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Description 给你n个点,m条无向边,每条边都有长度d和花费p,给你起点s终点t,要求输出起点到终点的最短距离及其花费,如果最短距离有多条路线,则输出花费最少的. Input 输入n,m,点的编号是1~n,然后是m行,每行4个数 a,b,d,p,表示a和b之间有一条边,且其长度为d,花费为p.最后一行是…

上周我们介绍了神奇的只有五行的 Floyd 最短路算法,它可以方便的求得任意两点的最短路径,这称为"多源最短路".本周来来介绍指定一个点(源点)到其余各个顶点的最短路径,也叫做"单源最短路径".例如求下图中的 1 号顶点到 2.3.4.5.6 号顶点的最短路径. 与 Floyd-Warshall 算法一样这里仍然使用二维数组 e 来存储顶点之间边的关系,初始值如下. 我们还需要用一个一维数组 dis 来存储 1 号顶点到其余各个顶点的初始路程,如下. 我们将此时 d…