http://poj.org/problem?id=1751

Description

The island nation of Flatopia is perfectly flat. Unfortunately, Flatopia has a very poor system of public highways. The Flatopian government is aware of this problem and has already constructed a number of highways connecting some of the most important towns. However, there are still some towns that you can't reach via a highway. It is necessary to build more 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 and town i has a position given by the Cartesian coordinates (xi, yi). Each highway connects exaclty two towns. All highways (both the original ones and the ones that are to be built) follow straight lines, and thus their length is equal to Cartesian distance between towns. 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 cost of building new highways. However, they want to guarantee that every town is highway-reachable from every other town. Since Flatopia is so flat, the cost of a highway is always proportional to its length. Thus, the least expensive highway system will be the one that minimizes the total highways length. 

Input

The input consists of two parts. The first part describes all towns in the country, and the second part describes all of the highways that have already been built. 
The first line of the input file contains a single integer N (1 <= N <= 750), representing the number of towns. The next N lines each contain two integers, xi and yi separated by a space. These values give the coordinates of ith town (for i from 1 to N). Coordinates will have an absolute value no greater than 10000. Every town has a unique location. 
The next line contains a single integer M (0 <= M <= 1000), representing the number of existing highways. The next M lines each contain a pair of integers separated by a space. These two integers give a pair of town numbers which are already connected by a highway. Each pair of towns is connected by at most one highway. 

Output

Write to the output a single line for each new highway that should be built in order to connect all towns with minimal possible total length of new highways. Each highway should be presented by printing town numbers that this highway connects, separated by a space. 
If no new highways need to be built (all towns are already connected), then the output file should be created but it should be empty. 

Sample Input


Sample Output


题意: 

有一个N个城市的无向图,给你N个城市的坐标,然后现在该无向图已经有M条边了,问你还需要添加总长为多少的边能使得该无向图连通.输出需要添加边的两端点编号即可.

思路:

本题就是求最小生成树的,但是由于本题不需要输出最终生成树的权值,那么我们在求两点距离的时候时间保存距离 dist=(x1-x2)*(x1-x2)+(y1-y2)*(y1-y2);即可,不用sqrt开方(因为开方费时间).

然后对于已经连接上的边,我们令这些边长为0,并添加到无向图中去即可(或令他们属于同一个并查集也行)

 #include <stdio.h>
#include <string.h>
#include <iostream>
#include <string>
#include <math.h>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <sstream>
const int INF=0x3f3f3f3f;
typedef long long LL;
const int mod=1e9+;
//const double PI=acos(-1);
#define Bug cout<<"---------------------"<<endl
const int maxn=;
using namespace std; struct edge_node
{
int to;
int val;
int next;
}Edge[maxn*maxn];
int Head[maxn];
int tot; struct point_node
{
int x;
int y;
}PT[]; void Add_Edge(int u,int v,double w)
{
Edge[tot].to=v;
Edge[tot].val=w;
Edge[tot].next=Head[u];
Head[u]=tot++;
} int lowval[maxn];
int pre[maxn];//记录每个点的双亲是谁 void Prim(int n,int st)//n为顶点的个数,st为最小生成树的开始顶点
{
fill(lowval+,lowval++n,INF);//不能用memset(lowval,INF,sizeof(lowval))
memset(pre,,sizeof(pre));
lowval[st]=-;
pre[st]=-;
for(int i=Head[st];i!=-;i=Edge[i].next)
{
int v=Edge[i].to;
int w=Edge[i].val;
lowval[v]=min(lowval[v],w);
pre[v]=st;
}
for(int i=;i<n-;i++)
{
int MIN=INF;
int k;
for(int i=;i<=n;i++)//根据编号从0或是1开始,改i从0--n-1和1--n
{
if(lowval[i]!=-&&lowval[i]<MIN)
{
MIN=lowval[i];
k=i;
}
}
if(MIN!=)//权值不为0,说明要修路
printf("%d %d\n",pre[k],k);
lowval[k]=-;
for(int j=Head[k];j!=-;j=Edge[j].next)
{
int v=Edge[j].to;
int w=Edge[j].val;
if(w<lowval[v])
{
lowval[v]=w;
pre[v]=k;
}
}
}
} int main()
{
int n,m;
scanf("%d",&n);
memset(Head,-,sizeof(Head));
tot=;
for(int i=;i<=n;i++)
{
scanf("%d %d",&PT[i].x,&PT[i].y);
}
for(int i=;i<=n;i++)
{
for(int j=i+;j<=n;j++)
{
int x,y;
x=PT[i].x-PT[j].x;
y=PT[i].y-PT[j].y;
int val=x*x+y*y;
Add_Edge(i,j,val);
Add_Edge(j,i,val);
}
}
scanf("%d",&m);
for(int i=;i<m;i++)
{
int u,v;
scanf("%d %d",&u,&v);
Add_Edge(u,v,);
Add_Edge(v,u,);
}
Prim(n,);
return ;
}

POJ-1751 Highways(最小生成树消边+输出边)的更多相关文章

  1. POJ 1751 Highways (最小生成树)

    Highways Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u Submit Sta ...

  2. POJ 1751 Highways(最小生成树Prim普里姆,输出边)

    题目链接:点击打开链接 Description The island nation of Flatopia is perfectly flat. Unfortunately, Flatopia has ...

  3. POJ 1751 Highways (最小生成树)

    Highways 题目链接: http://acm.hust.edu.cn/vjudge/contest/124434#problem/G Description The island nation ...

  4. POJ 1751 Highways 【最小生成树 Kruskal】

    Highways Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 23070   Accepted: 6760   Speci ...

  5. POJ 2485 Highways(最小生成树+ 输出该最小生成树里的最长的边权)

                                                                                                         ...

  6. POJ 1751 Highways(最小生成树&Prim)题解

    思路: 一开始用Kruskal超时了,因为这是一个稠密图,边的数量最惨可能N^2,改用Prim. Prim是这样的,先选一个点(这里选1)作为集合A的起始元素,然后其他点为集合B的元素,我们要做的就是 ...

  7. POJ 1751 Highways (kruskal)

    题目链接:http://poj.org/problem?id=1751 题意是给你n个点的坐标,然后给你m对点是已经相连的,问你还需要连接哪几对点,使这个图为最小生成树. 这里用kruskal不会超时 ...

  8. POJ 1751 Highways (ZOJ 2048 ) MST

    http://poj.org/problem?id=1751 http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2048 题目大 ...

  9. (poj) 1751 Highways

    Description The island nation of Flatopia is perfectly flat. Unfortunately, Flatopia has a very poor ...

随机推荐

  1. Codeforces 392 B Blown Garland

    题意:输入是由连续的RYGB和字符!组成的字符串,R代表红灯,Y代表黄灯,G代表绿灯,B代表蓝灯.简而言之,就是给定的字符串保证了下标对4取余相同的灯颜色都相同,但是有的地方为‘!’代表这个灯坏了,然 ...

  2. PyTorch实战:经典模型LeNet5实现手写体识别

    在上一篇博客CNN核心概念理解中,我们以LeNet为例介绍了CNN的重要概念.在这篇博客中,我们将利用著名深度学习框架PyTorch实现LeNet5,并且利用它实现手写体字母的识别.训练数据采用经典的 ...

  3. 三十一、CI框架之使用验证码

    一.CI的验证码功能用着很是舒服,需要在根目录下新建一个captcha的验证码文件夹用于存放生产的图片,代码如下: 二.浏览器效果如下: 总结:关于验证码生产函数,有很多参数可以设置,包括字体,验证码 ...

  4. Mac下使用Hexo搭建个人博客

    Hexo介绍 利用原作者的一句话:A fast,simple&powerful blog framework,powered by Node.js Hexo是基于Node.js的博客平台,He ...

  5. 学习spring的第二天

    对昨天的查漏:关于<bean>标签的scope属性,是由它决定原型和单例的,而不是说你java代码中用到了单例模式就是单例了. 其二就是lazy-init属性,它对于scope=" ...

  6. part11 Vue项目接口联调//真机测试

    何为项目接口联调? 前端代码编译好了  后端接口写好了 我们就需要去掉前端模拟数据干掉 用后端提供的数据.进行前后端的一个调试 如何联调? config目录下面 index.js 文件 dev 中pr ...

  7. scala通过尾递归解析提取字段信息

    一.背景 获取数据中以“|”作为字段间的分隔符,但个别字段中数据也是以“|”作为分隔符.因此,在字段提取时需要保护数据完整性. 二.实现 1.数据以“|”分隔,可以采用递归方式迭代解析.通过尾递归方式 ...

  8. 洛谷 P1709 隐藏口令

    题目描述 有时候程序员有很奇怪的方法来隐藏他们的口令.Binny会选择一个字符串S(由N个小写字母组成,5<=N<=5,000,000),然后他把S顺时针绕成一个圈,每次取一个做开头字母并 ...

  9. .net core excel导入导出

    做的上一个项目用的是vs2013,传统的 Mvc模式开发的,excel报表的导入导出都是那几段代码,已经习惯了. 导入:string filename = ExcelFileUpload.FileNa ...

  10. scrapy练习1

    1.建立项目: #建立名为tuto的项目 scrapy startproject tuto 2.进入项目目录: cd tuto 3.建立域名任务: #minyan任务名:后面是任务对应的域名 scra ...