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…

There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a…

Constructing Roads Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 19884   Accepted: 8315 Description There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each…

Constructing Roads Description There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road betw…

Constructing Roads 题目链接: http://acm.hust.edu.cn/vjudge/contest/124434#problem/D Description There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two villa…

题意:给出n个村庄之间的距离,再给出已经连通起来了的村庄.求把所有的村庄都连通要修路的长度的最小值. 思路:Kruskal算法 课本代码: //Kruskal算法 #include<iostream> using namespace std; int fa[120]; int get_father(int x){ return fa[x]=fa[x]==x?x:get_father(fa[x]);//判断两个节点是否属于一颗子树(并查集) } int main(){ int n; int p[…

There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a…

Description There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or the…

给一个n个点的完全图 再给你m条道路已经修好 问你还需要修多长的路才能让所有村子互通 将给的m个点的路重新加权值为零的边到边集里 然后求最小生成树 #include<cstdio> #include<iostream> #include<algorithm> #include<cmath> #include<cstring> #include<string> #define cl(a,b) memset(a,b,sizeof(a))…

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[…