Description

The island nation of Flatopia is perfectly flat. Unfortunately, Flatopia has no public highways. So the traffic is difficult in Flatopia. The Flatopian government is aware of this problem. They're planning to build some highways so that it will be possible
to drive between any pair of towns without leaving the highway system. 



Flatopian towns are numbered from 1 to N. Each highway connects exactly two towns. All highways follow straight lines. All highways can be used in both directions. Highways can freely cross each other, but a driver can only switch between highways at a town
that is located at the end of both highways. 



The Flatopian government wants to minimize the length of the longest highway to be built. However, they want to guarantee that every town is highway-reachable from every other town.

Input

The first line of input is an integer T, which tells how many test cases followed. 

The first line of each case is an integer N (3 <= N <= 500), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 65536]) between
village i and village j. There is an empty line after each test case.

Output

For each test case, you should output a line contains an integer, which is the length of the longest road to be built such that all the villages are connected, and this value is minimum.

Sample Input

1

3
0 990 692
990 0 179
692 179 0

Sample Output

692

#include<stdio.h>
#include<string.h>
#define INF 0x3f3f3f3f
#define N 550
int n;
int i,j;
int map[N][N];
int a,b;
int low[N];
bool vis[N];
int sum[N];
void input()
{
//memset(map,INF,sizeof(map));
for(i=1;i<=n;++i)
{
for(j=1;j<=n;++j)
{
scanf("%d",map[i]+j);
}
}
}
void prim()
{
int pos=1;
for(i=1;i<=n;++i)//第一次给low赋值
{
low[i]=map[pos][i];
}
vis[pos]=1; //增加最小生成树集合
for(i=1;i<n;++i)//再找n-1个点
{
int min=INF;
for(j=1;j<=n;++j)
{
if(!vis[j]&&min>low[j])
{
min=low[j];
pos=j;//把找到的点记录下
}
}
sum[i]=min;
vis[pos]=1;
for(j=1;j<=n;++j)
{
if(!vis[j]&&low[j]>map[pos][j])
{
low[j]=map[pos][j];
}
}
}
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
memset(vis,0,sizeof(vis));
input();
prim();
int max=-INF;
for(i=1;i<n;++i)
if(max<sum[i])
max=sum[i];
printf("%d\n",max);
}
return 0;
}

Highways POJ 2485【Prim】的更多相关文章

  1. poj2728 Desert King【最优比率生成树】【Prim】【0/1分数规划】

    含[最小生成树Prim]模板. Prim复杂度为$O(n^2),适用于稠密图,特别是完全图的最小生成树的求解.   Desert King Time Limit: 3000MS   Memory Li ...

  2. 【简●解】POJ 1845 【Sumdiv】

    POJ 1845 [Sumdiv] [题目大意] 给定\(A\)和\(B\),求\(A^B\)的所有约数之和,对\(9901\)取模. (对于全部数据,\(0<= A <= B <= ...

  3. Highways - poj 2485 (Prim 算法)

      Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 24383   Accepted: 11243 Description T ...

  4. 洛谷1546 最短网络Agri-Net【最小生成树】【prim】

    [内含最小生成树Prim模板] 题目:https://www.luogu.org/problemnew/show/P1546 题意:给定一个邻接矩阵.求最小生成树. 思路:点少边多用Prim. Pri ...

  5. POJ 2154 【POLYA】【欧拉】

    前记: TM终于决定以后干啥了.这几天睡的有点多.困饿交加之间喝了好多水.可能是灌脑了. 切记两件事: 1.安心当单身狗 2.顺心码代码 题意: 给你N种颜色的珠子,串一串长度问N的项链,要求旋转之后 ...

  6. Highways poj 2485

    Description The island nation of Flatopia is perfectly flat. Unfortunately, Flatopia has no public h ...

  7. POJ 3230 【DP】

    题意: 某货旅行,在n个城市呆m天. 给出从第i个城市到第j个城市的路费,或者留在某个城市的生活费. 给出在第i天在第j个城市的收益. 可以在城市之间任意穿梭逗留没有其他特殊要求. 求收益最大是多少. ...

  8. H - Highways - poj 1751(prim)

    某个地方政府想修建一些高速公路使他们每个乡镇都可以相同通达,不过以前已经修建过一些公路,现在要实现所有的联通,所花费的最小代价是多少?(也就是最小的修建长度),输出的是需要修的路,不过如果不需要修建就 ...

  9. POJ 3187【permutation】

    POJ 3187 给定N值,从而确定了数据的范围及长度,暴力枚举数列,接下来类似杨辉三角的递推计算.注permutation从递增有序数列开始枚举,枚举到符合sum值时退出即可 #include &l ...

随机推荐

  1. redis基础一_常用指令

    # Redis configuration file example. # # Note that in order to read the configuration file, Redis mus ...

  2. Archive for required library: 'D:/Program Files/Apache/maven-repository/dom4j/dom4j/1.6.1/dom4j-1.6.1.jar'

    今天导入一个项目工程,发现报错:Archive for required library: 'D:/Program Files/Apache/maven-repository/dom4j/dom4j/ ...

  3. 前端 (cookie )页面进入存储一次

     <!--引入jq--> <script> var isShowTip = window.sessionStorage.getItem("isShow") ...

  4. HDU - 5438 Ponds(拓扑排序删点+并查集判断连通分量)

    题目: 给出一个无向图,将图中度数小于等于1的点删掉,并删掉与他相连的点,直到不能在删为止,然后判断图中的各个连通分量,如果这个连通分量里边的点的个数是奇数,就把这些点的权值求和. 思路: 先用拓扑排 ...

  5. C/C++连接MySQL数据库执行查询

    1. 简介: 使用C/C++连接MySQL数据库执行增删改查操作,基本就是围绕以下两个文件展开: mysql.h(此头文件一般在MySQL的include文件夹内,如 D:\MySQL\mysql-5 ...

  6. python TCP协议详解 三次握手四次挥手和11种状态

    11种状态解析 LISTEN  --------------------  等待从任何远端TCP 和端口的连接请求. SYN_SENT  ---------------  发送完一个连接请求后等待一个 ...

  7. Android设置透明状态

    xml中: android:background="@android:color/transparent" 半透明: android:background="#e0000 ...

  8. 算法导论 第八章 线性时间排序(python)

    比较排序:各元素的次序依赖于它们之间的比较{插入排序O(n**2) 归并排序O(nlgn) 堆排序O(nlgn)快速排序O(n**2)平均O(nlgn)} 本章主要介绍几个线性时间排序:(运算排序非比 ...

  9. Android开发——Accessibility机制实现模拟点击(微信自动抢红包实现)

    1. 何为Accessibility机制 许多Android使用者因为各种情况导致他们要以不同的方式与手机交互.对于那些由于视力.听力或其它身体原因导致不能方便使用Android智能手机的用户,And ...

  10. URAL 2040 Palindromes and Super Abilities 2

    Palindromes and Super Abilities 2Time Limit: 500MS Memory Limit: 102400KB 64bit IO Format: %I64d &am ...