hdu 3371(kruskal)

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Connect the Cities Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 15569 Accepted Submission(s): 4108 Problem Description In 2100, since the sea level rise, most of the cities disappear. Thou…

HDU 3371 Connect the Cities(并查集+Kruskal)

题目网址:http://acm.hdu.edu.cn/showproblem.php?pid=3371 思路: 这道题很明显是一道最小生成树的题目,有点意思的是,它事先已经让几个点联通了.正是因为它先联通了几个点,所以为了判断连通性很容易想到用并查集+kruskal. 不过要注意这题有一个坑点,就是边数很多上限是25000,排序的话可能就超时了.而点数则比较少上限是500,所以很多人选择用Prim做.但我个人觉得这样连通性不好判断.其实边数多没关系,我们只要去重就好啦,用邻接矩阵存下两点…

HDU 3371(城市联通最小生成树-Kruskal)

题意是求将所有点联通所花费的最小金额,如不能完全联通,输出 -1 直接Kruskal,本题带来的一点教训是 rank 是algorithm头文件里的,直接做变量名会导致编译错误.没查到 rank 的具体用途...... #include <cstdio> #include <iostream> #include <algorithm> /*rank 是algorithm里的*/ using namespace std; ],r[]; int n,m,k,ans; str…

hdu 3371 Connect the Cities

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=3371 Connect the Cities Description In 2100, since the sea level rise, most of the cities disappear. Though some survived cities are still connected with others, but most of them become disconnected. The…

hdu 3371 Connect the Cities (最小生成树Prim)

题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=3371 题目不难稍微注意一下要把已经建好的城市之间的花费定义为0,在用普通Prim算法就可以了:我没有用克鲁斯卡尔算法(Kruskal's algorithm),因为这题数据比较大,而且要处理大量的数据使它为0,怕超时T^T..... #include<iostream> #include<cstdio> #include<algorithm> using namespa…

Hdu 3371 Connect the Cities（最小生成树）

地址:http://acm.hdu.edu.cn/showproblem.php?pid=3371 其实就是最小生成树,但是这其中有值得注意的地方:就是重边.题目没有告诉你两个城市之间只有一条路可走,所以两个城市之间可能有多条路可以走. 举例: 输入可以包含 1 2 3 // 1到2的成本为3 1 2 5 //1到2的成本为5 因此此时应选成本较低的路. 然后,已经连通的城市之间的连通成本为0. 这题用G++提交得到984ms的反馈,用C++提交则得到484ms的反馈. 很想知…

HDU 3371 Connect the Cities（prim算法）

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=3371 Problem Description In 2100, since the sea level rise, most of the cities disappear. Though some survived cities are still connected with others, but most of them become disconnected. The governmen…