Minimum Transport Cost http://acm.hdu.edu.cn/showproblem.php?pid=1385 Problem Description These are N cities in Spring country. Between each pair of cities there may be one transportation track or none. Now there is some cargo that should be delivere…

Minimum Transport Cost Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 8794    Accepted Submission(s): 2311 Problem Description These are N cities in Spring country. Between each pair of cities…

Minimum Transport CostTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 12860    Accepted Submission(s): 3633 Problem DescriptionThese are N cities in Spring country. Between each pair of cities t…

题意:给你n个城市,一些城市之间会有一些道路,有边权.并且每个城市都会有一些费用. 然后你一些起点和终点,问你从起点到终点最少需要多少路途. 除了起点和终点,最短路的图中的每个城市的费用都要加上. 思路一:因为有多组数据,所以可以采用弗洛伊德算法求多源最短路. 但是,这里要求输出的路径,且按字典序输出. 这里可以用一个数组:pre[i][j]表示i到j的路径上的首个付费城市.这是最关键的地方. 要注意:输出时候,如果起点和终点相同.只输出i,没有箭头. #include <iostream>…

<题目链接> 题目大意:给你一张图,有n个点,每个点都有需要缴的税,两个直接相连点之间的道路也有需要花费的费用.现在进行多次询问,给定起点和终点,输出给定起点和终点之间最少花费是多少,并且输出最少花费所走的路径,如果有多条路径花费最少,则输出字典序最小的那条. 解题分析: 输出最短路的路径问题,需要注意的是,题目要求输出的最短路径的字典序最小,所以我们在每次松弛的时候,都需要加上判断.如果有多个点的最短路相同,则用DFS求出它们之前走过的路径,并且进行比较,然后选字典序最小的那条. #incl…

貌似···················· 这个算法深的东西还是很不熟悉!继续学习!!!! ++++++++++++++++++++++++++++ ============================ ++++++++++++++++++++++++++++ ------------------------------------------------- ============================ #include<stdio.h> #include<string…

http://acm.hdu.edu.cn/showproblem.php?pid=1385 #include <cstdio> #include <cstring> #include <algorithm> #define maxn 1001 using namespace std; ; int g[maxn][maxn],tax[maxn],pre[maxn][maxn]; int m,n,s,e; void inti() { ; i<=n; i++) { ;…

链接:传送门 题意:有 n 个城市,从城市 i 到城市 j 需要话费 Aij ,当穿越城市 i 的时候还需要话费额外的 Bi ( 起点终点两个城市不算穿越 ),给出 n × n 大小的城市关系图,-1 代表两个城市无法连通,询问若干次,求出每次询问的两个城市间的最少花费以及打印出路线图 思路:经典最短路打印路径问题,直接使用一个二维数组 path[i][j] 记录由点 i 到点 j 最短路的最后后继点,这道题关键是当松弛操作相等时,i 到 j 的最短距离是可以由 k 进行跳转的,这时候需要判断这…

本题就是使用Floyd算法求全部路径的最短路径,并且须要保存路径,并且更进一步须要依照字典顺序输出结果. 还是有一定难度的. Floyd有一种非常巧妙的记录数据的方法,大多都是使用这种方法记录数据的. 只是事实上本题数据不是非常大,一般太大的数据也无法使用Floyd,由于效率是O(N^3). 所以事实上也能够使用一般的Floyd算法,然后添加个三维数组记录数据.以下就是这样的做法,0ms过了. #include <stdio.h> #include <vector> using s…

Floyd.注意字典序!!! #include <stdio.h> #include <string.h> #define MAXNUM 55 #define INF 0x1fffffff int cost[MAXNUM][MAXNUM]; int path[MAXNUM][MAXNUM]; int taxes[MAXNUM]; int que[MAXNUM]; int n; void floyd(int n) { int i, j, k, tmp; ; i<=n; ++i)…

题目链接: huangjing 思路: 输出路径的最短路变种问题..这个题目在于多组询问.那么个人认为用floyd更加稳妥一点.还有就是在每一个城市都有过路费,所以在floyd的时候更改一下松弛条件就可以..那么输出路径怎么办呢??我採用的是输出起点的后继而不是终点的前驱..由于我们关心的是路径字典序最小,关心的是起点的后继...那么打印路径的时候就直接从前向后打印,这个和dijkstra的打印路径稍有不同... 最短路的打印參见传送门 题目: Minimum Transport Cost Ti…

Minimum Transport Cost Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8860    Accepted Submission(s): 2331 Problem Description These are N cities in Spring country. Between each pair of cities…

Minimum Transport Cost Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 10029    Accepted Submission(s): 2716 Problem Description These are N cities in Spring country. Between each pair of cities…

http://acm.hdu.edu.cn/showproblem.php? pid=1385 求最短路.要求输出字典序最小的路径. spfa:拿一个pre[]记录前驱,不同的是在松弛的时候.要考虑和当前点的dis值相等的情况,解决的办法是dfs找出两条路径中字典序较小的.pre[]去更新. 把路径当做字符串处理. 我仅仅用之前的pre去更新当前点,并没考虑到起点到当前点的整个路径,事实上这样并不能保证是字典序最小.wa了N次.于是乎搜了下题解,发现用spfa解的非常少.看到了某大牛的解法如上,…

These are N cities in Spring country. Between each pair of cities there may be one transportation track or none. Now there is some cargo that should be delivered from one city to another. The transportation fee consists of two parts: The cost of the…

These are N cities in Spring country. Between each pair of cities there may be one transportation track or none. Now there is some cargo that should be delivered from one city to another. The transportation fee consists of two parts: The cost of the…

题目链接: https://vjudge.net/problem/ZOJ-1456 These are N cities in Spring country. Between each pair of cities there may be one transportation track or none. Now there is some cargo that should be delivered from one city to another. The transportation f…

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1456 题意:求最短路并且输出字典序最小的答案. 思路:如果用dijkstra来做的话,会比较麻烦,这里直接用floyd会简单的多,只需要记录好后继路径即可. #include<iostream> #include<algorithm> #include<cstring> #include<cstdio> #include<sst…

最短路的路径打印问题 同时路径要是最小字典序 字典序用floyd方便很多 学会了两种打印路径的方法!!! #include <stdio.h> #include <string.h> #define N 110 #define INF 1000000000 int d[N][N],path[N][N],c[N]; int n,cost; int s,t; void input() { int i,j,w; ; i<=n; i++) ; j<=n; j++) { scan…

求最短路的算法最有名的是Dijkstra.所以一般拿到题目第一反应就是使用Dijkstra算法.但是此题要求的好几对起点和终点的最短路径.所以用Floyd是最好的选择.因为其他三种最短路的算法都是单源的. 输出字典序最小的路径则需要修改模版. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; , INF=; int Ma…

题目大意 求多组i到j的最短路径 并输出字典序最小.... 现在只会floyd的方式 利用dis[i][j] 表示i到j的路径中i 后面的节点 更新是比较dis[i][j] dis[i][k]. 记住这个就好 ,其余存法貌似会有问题.代码如下: #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <ctime> #include <…

Ignatius and the Princess I Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 10312    Accepted Submission(s): 3125 Special Judge Problem Description The Princess has been abducted by the BEelzeb…

Minimum Transport Cost Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 10183    Accepted Submission(s): 2791 Problem Description These are N cities in Spring country. Between each pair of cities…

Minimum Transport Cost Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9109    Accepted Submission(s): 2405 Problem Description These are N cities in Spring country. Between each pair of cities…

Minimum Transport Cost Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 11017    Accepted Submission(s): 3058 Problem Description These are N cities in Spring country. Between each pair of cities…

Minimum Transport Cost Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9814    Accepted Submission(s): 2641 Problem Description These are N cities in Spring country. Between each pair of cities…

Given an integer array, adjust each integers so that the difference of every adjacent integers are not greater than a given number target. If the array before adjustment is A, the array after adjustment is B, you should minimize the sum of |A[i]-B[i]…

Given an integer array, adjust each integers so that the difference of every adjcent integers are not greater than a given number target. If the array before adjustment is A, the array after adjustment is B, you should minimize the sum of |A[i]-B[i]|…

HDU 4280 Island Transport(网络流,最大流) Description In the vast waters far far away, there are many islands. People are living on the islands, and all the transport among the islands relies on the ships. You have a transportation company there. Some route…

HDU.1394 Minimum Inversion Number (线段树 单点更新 区间求和 逆序对) 题意分析 给出n个数的序列,a1,a2,a3--an,ai∈[0,n-1],求环序列中逆序对最少的个数. 前置技能 环序列 还 线段树的逆序对求法 逆序对:ai > aj 且 i < j ,换句话说数字大的反而排到前面(相对后面的小数字而言) 环序列:把第一个放到最后一个数后面,就是一次成环,一个含有n个元素序列有n个环序列. 线段树的逆序对求法:每个叶子节点保存的是当前值数字的个数.根…