http://poj.org/problem?id=2421 这个题和poj1258是一样的,只要在1258的基础上那么几行代码,就可以A,水. 题意:还是n连通问题,和1258不同的就是这个还有几条路在之前就已经连通了的,所以不需要再去连. #include <stdio.h> #include <string.h> #define inf 100009 ]; ][],dis[],ans,n; int prim() { ;i<=n;i++) dis[i]=inf;dis[]…

poj 1251  && hdu 1301 Sample Input 9 //n 结点数A 2 B 12 I 25B 3 C 10 H 40 I 8C 2 D 18 G 55D 1 E 44E 2 F 60 G 38F 0G 1 H 35H 1 I 353A 2 B 10 C 40B 1 C 200Sample Output 21630 prim算法 # include <iostream> # include <cstdio> # include <cstr…

Constructing Roads Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 2421 Appoint description:  System Crawler  (2015-05-27) Description There are N villages, which are numbered from 1 to N, and y…

题目链接:http://poj.org/problem?id=2421 最小生成树的变形,有的村庄已经连接了,就直接把他们的权值赋为0,一样的做最小生成树,Prim算法. #include <stdio.h> #include <string.h> #define INF 0x3f3f3f3f ][]; ]; ]; int n; int Prim() { memset(vis,false,sizeof(vis)); ;i<=n;i++) dis[i] = INF; ; dis…

题意:有几个村庄,要修最短的路,使得这几个村庄连通.但是现在已经有了几条路,求在已有路径上还要修至少多长的路. 分析:用Prim求最小生成树,将已有路径的长度置为0,由于0是最小的长度,所以一定会被Prim选中加入最小生成树. package Map; import java.util.Scanner; /** * Prime */ public class Poj_2421_Prim { static int MAXVEX = 200; static int n, m; static int[…

链接: http://poj.org/problem?id=2421 Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 21130   Accepted: 8924 Description There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can…

题目链接:http://poj.org/problem?id=1251 字符用%s好了,方便一点. #include <stdio.h> #include <string.h> #define INF 0x3f3f3f3f ][]; ]; ]; int n; int Prim() { memset(vis,false,sizeof(vis)); ; i<=n; i++) dis[i] = INF; ; dis[] = ; ; i<=n; i++) { ; ; j<…

题目链接:http://poj.org/problem?id=2421 实际上又是考最小生成树的内容,也是用到kruskal算法.但稍稍有点不同的是,给出一些已连接的边,要在这些边存在的情况下,拓展出最小生成树来. 一般来说,过到这四组数据大体上就能AC了.  1.题目给出的案例数据    2.连接的道路可能把所有的村庄都已经连通了       3.两个村庄给出多次,即连接这两个村庄的道路是重复的!. 2.3这两种情况的数据如下(为了好看,自己出了一组,当然Sample Input 那组也行):…

Building a Space Station Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 4400 Accepted: 2255 Description You are a member of the space station engineering team, and are assigned a task in the construction process of the station. You are ex…

题意:要在n个城市之间建造公路,使城市之间能互相联通,告诉每个城市之间建公路的费用,和已经建好的公路,求最小费用. 解法:最小生成树.先把已经建好的边加进去再跑kruskal或者prim什么的. 代码: #include<stdio.h> #include<iostream> #include<algorithm> #include<string> #include<string.h> #include<math.h> #includ…