Agri-Net Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 64912   Accepted: 26854 Description Farmer John has been elected mayor of his town! One of his campaign promises was to bring internet connectivity to all farms in the area. He nee…

Agri-Net Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 45050   Accepted: 18479 Description Farmer John has been elected mayor of his town! One of his campaign promises was to bring internet connectivity to all farms in the area. He nee…

题意:给全村建光纤,求花费最小 思路:最小生成树,树相对于图来说就是没有环 m用来存图 v判断是否访问 low用来存两点间的最短距离 给low赋值  for(i=1;i<=n;i++){if(i!=pos)  low[i]=m[pos][i]  else low[i]=0;} 找最小值 即找顶点i的最小距离  for(int j=1;j<n;j++)  for(int i=1;i<=n;i++)  if(v[i]==0&&min>l[i]) min=l[i] pos…

题面传送门 开始挖老祖宗(ycx)留下来的东西.jpg 本来想水一道紫题作为 AC 的第 500 道紫题的,结果发现点开了道神题. 首先先讲一个我想出来的暴力做法.条件一和条件二直接扫一遍判断掉.先将所有点按照 \(a_{i,j}\) 按权值大小从小到大排序并依次插入这些点,我们实时维护一个 \(n\times n\) 的 bool 数组 \(vis\),\(vis_{i,j}\) 表示第 \(i\) 行第 \(j\) 列的数是否被访问了.当我们插入某个 \(a_{i,j}\) 时,如果 \(\…

题意:n个农场,求把所有农场连接起来所需要最短的距离. 思路:prim算法 课本代码: //prim算法 #include<iostream> #include<stdio.h> #include<cstring> using namespace std; int n; int tot; int v[150][150]; int dist[150];//存 节点到树 的最小距离 bool use[150];//标记节点是否存在 int main(){ while(sca…

(￣▽￣)" #include<iostream> #include<cstdio> #include<cmath> #include<algorithm> #include<cstring> #include<string> #include<cstdlib> #include<vector> using namespace std; typedef long long ll; const int…

一直以来只会Kruskal prim和dijkstra很像 只不过prim维护的是最短的边,而dijkstra维护的是最短的从起点到一个点的路径 同时prim要注意当前拓展的边是没有拓展过的 可以用堆优化 #include<bits/stdc++.h> #define REP(i, a, b) for(register int i = (a); i < (b); i++) #define _for(i, a, b) for(register int i = (a); i <= (b…

这个时间短 700多s #include<stdio.h> #include<string.h> #include<iostream> #include<algorithm> using namespace std; struct node{ int u; int v; int w; }que[100000]; int father[505]; bool cmp(struct node a,struct node b){ return a.w<b.w;…

Agri-Net Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 20000/10000K (Java/Other) Total Submission(s) : 1   Accepted Submission(s) : 1 Problem Description Farmer John has been elected mayor of his town! One of his campaign promises was to bri…

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[…