题意,在给出的图中,使用最小花费的边,使这个图仍然连通. #include <cstdio> #include <algorithm> #include <cstring> using namespace std; const int maxn=10005; int head[maxn]; int n,len=0,counter; long long ans; struct node{ int v,cost,u; //操作符的重写默认是小于等于 bool operato…

1.TreeMap和TreeSet类:A - Language of FatMouse ZOJ1109B - For Fans of Statistics URAL 1613 C - Hardwood Species POJ 2418D - StationE - Web Navigation ZOJ 1061F - Argus ZOJ 2212G - Plug-in2.SegmentTreeA-敌兵布阵 hdu 1166B - I Hate It HDU 1754C - A Simple Pro…

题目链接:poj1258 Agri-Net 这题我上个月做过,是个大水题,今天看见有人用prim+heap做的,就学习了下. #include<cstdio> #include<cstring> #include<algorithm> #include<queue> #include<vector> #include<set> #define CLR(a,b) memset((a),(b),sizeof((a))) using nam…

题目链接: https://vjudge.net/problem/POJ-1258 题目大意: 求MST 思路: 由于给的是邻接矩阵,直接prim算法 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<queue> #include<stack> #include<map…

题目链接:http://poj.org/problem?id=1258 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 needs your help, of course. Farmer John ordered a high speed…

题目链接 http://poj.org/problem?id=1258 题意 有n个农场,现在要在n个农场之间铺设光纤使得n个农场连接起来,求铺设光纤的最短距离. 思路 最小生成树问题,使用Prime算法或者Kruskal算法解决. 代码 Prime算法: #include <algorithm> #include <iostream> #include <cstring> #include <cstdio> using namespace std; con…

题意:给你一个矩阵M[i][j]表示i到j的距离 求最小生成树 思路:裸最小生成树 prime就可以了 最小生成树专题 AC代码: #include "iostream" #include "string.h" #include "stack" #include "queue" #include "string" #include "vector" #include "set&…

Agri-Net Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 44032   Accepted: 18001 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…

http://poj.org/problem?id=1258 这道题是最简单的一个啦,,,, #include<stdio.h> #include<iostream> #include<string.h> #include<algorithm> using namespace std; ; #define M 0x3f3f3f3f///设置无穷大的数M,当两点之间无法联通时设两者之间距离为M int ma[N][N],low[N],vis[N];///ma记…

本文出自:http://blog.csdn.net/svitter 题意:给出一个数字n代表邻接矩阵的大小,随后给出邻接矩阵的值.输出最小生成树的权值. 题解: prime算法的基本解法: 1.选择一个点,然后不停的向当中增加权值最小的边,边的一端在已经生成的部分生成树中,还有一端在未生成的生成树中. 2.利用优先队列维护边,将增加的点所包括的边增加到队列中去,随后依照边的权值弹出. 简单理解方法:一个人能够拉非常多人,新被拉进来的人,把能拉的人(有边,且未被訪问)排队,找最好拉的人拉进来,循环…