Jungle Roads Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6211    Accepted Submission(s): 4512 Problem Description The Head Elder of the tropical island of Lagrishan has a problem. A burst of…

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…

HDU 1233(最小生成树 模板) #include <iostream> #include <algorithm> #include <cstdio> using namespace std; typedef struct { int a,b; int v; }node; const int maxn=105; int ans; int father[maxn]; node graph[maxn*(maxn-1)/2]; int Find(int x) { if(f…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1301 The Head Elder of the tropical island of Lagrishan has a problem. A burst of foreign aid money was spent on extra roads between villages some years ago. But the jungle overtakes roads relentlessly,…

地址:http://acm.hdu.edu.cn/showproblem.php?pid=1301 很明显,这是一道“赤裸裸”的最小生成树的问题: 我这里采用了Kruskal算法,当然用Prim算法也一样可以解题. #include <iostream> #include <cstring> #include <cstdio> #include <cstdlib> using namespace std; typedef struct node{ int f…

链接: http://acm.hdu.edu.cn/showproblem.php?pid=1301 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=82831#problem/M Jungle Roads Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5741    Acce…

双向边,基础题,最小生成树   题目 同题目     #define _CRT_SECURE_NO_WARNINGS #include <stdio.h> #include<string.h> #include <malloc.h> #include<stdlib.h> #include<algorithm> #include<iostream> using namespace std; #define inf 999999999 #…

Jungle Roads 题目链接: http://acm.hust.edu.cn/vjudge/contest/124434#problem/A http://acm.hust.edu.cn/vjudge/contest/124434#problem/L Description The Head Elder of the tropical island of Lagrishan has a problem. A burst of foreign aid money was spent on e…

题解 这是一道裸的最小生成树题,拿来练手,题目就不放了 个人理解  Prim有些类似最短路和贪心,不断找距当前点最小距离的点 Kruskal类似于并查集,不断找最小的边,如果不是一棵树的节点就合并为一颗树 AC代码: Prim算法: #include<iostream> #include<cstdio> //EOF,NULL #include<cstring> //memset #include<cstdlib> //rand,srand,system,it…

裸的最小生成树 输入很蓝瘦 **并查集 int find(int x) { return x == fa[x] ? x : fa[x] = find(fa[x]); } 找到x在并查集里的根结点,如果两个端点在同一个集合内,find之后两个值就相等了 每次找到权值最小的端点不在同一集合的边 把两个集合合并 #include <iostream> #include <cstdio> #include <algorithm> using namespace std; ];…