题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1301 依旧Prim............不多说了 #include<iostream> #include<cstdio> #include<algorithm> using namespace std; #define MAX 'Z'+5 #define inf 99999999 int map[MAX][MAX],node[MAX],vis[MAX],n,path; v…

D - Jungle Roads Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 1251 Description The Head Elder of the tropical island of Lagrishan has a problem. A burst of foreign aid money was spent on extr…

Jungle Roads Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 16   Accepted Submission(s) : 12 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description #include <iostream> #in…

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…

Jungle Roads Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 175 Accepted Submission(s): 159   Problem Description The Head Elder of the tropical island of Lagrishan has a problem. A burst of fore…

题意  有n个村子  输入n  然后n-1行先输入村子的序号和与该村子相连的村子数t  后面依次输入t组s和tt s为村子序号 tt为与当前村子的距离  求链接全部村子的最短路径 还是裸的最小生成树咯 #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int N=30,M=1000; int par[N],n,m,ans; struct edge{int u…

题目: Description 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, so the large road network is too exp…

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…

c/c++ 用普利姆(prim)算法构造最小生成树 最小生成树(Minimum Cost Spanning Tree)的概念: ​ 假设要在n个城市之间建立公路,则连通n个城市只需要n-1条线路.这时,自然会考虑,如何在最节省经费的前提下建立这个公路网络. ​ 每2个城市之间都可以设置一条公路,相应地都要付出一定的经济代价.n个城市之间,最多可以设置n(n-1)/2条线路,那么,如何在这些可能的线路中选择n-1条,以使总的耗费最少? 普利姆(prim)算法的大致思路: ​ 大致思想是:设图G顶点…

HDU.1233 还是畅通工程(Prim) 题意分析 首先给出n,代表村庄的个数 然后出n*(n-1)/2个信息,每个信息包括村庄的起点,终点,距离, 要求求出最小生成树的权值之和. 注意村庄的编号从1开始即可 直接跑prim 代码总览 #include <bits/stdc++.h> #define nmax 105 #define inf 1e8+7 using namespace std; int mp[nmax][nmax]; int n; int totaldis =0; void…