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…

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…

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…

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…

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.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…

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…

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>…

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…

利用Floyd的DP状态转移方程. #include<cstdio> #include<cstring> #include<queue> #include<algorithm> using namespace std; #define MAXN 111 int n,m; double d[MAXN][MAXN]; void Floyd() { ; k<=n; ++k) { ; i<=n; ++i) { ; j<=n; ++j) { || d…