hdu 1301】的更多相关文章

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,…
http://acm.hdu.edu.cn/showproblem.php?pid=1301 #include <cstdio> #include <cstring> #include <algorithm> #define maxn 500 using namespace std; <<; int g[maxn][maxn]; int dis[maxn]; bool vis[maxn]; int n,x,mm,sum; char ch,ch1; void…
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…
最小生成树模板题 简单的prim算法 AC代码: #include <iostream> #include <stdio.h> #define INF 9999999 using namespace std; int map[200][200],dis[200],vis[200]; int prim(int n) { int i,j,min,p; for(i=1;i<=n;i++) { dis[i]=map[1][i]; vis[i]=0; } vis[1]=1; int s…
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…
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…
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; ];…