poj1751最小生成树
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 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 i th 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
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
9
1 5
0 0
3 2
4 5
5 1
0 4
5 2
1 2
5 3
3
1 3
9 7
1 2
Sample Output
1 6
3 7
4 9
5 7
8 3
此题纠结了好久,重点是想办法输出,把所有更新了的节点记录,到下一次更新时输出
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
using namespace std;
const double inf=1e20;
double d[800],cost[800][800];
int x[800],y[800];
int n,m;
struct node{
double x,y;
}e[800];
double dis(node a,node b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
void prim()
{
bool vis[800];
for(int i=1;i<=n;i++)
{
vis[i]=0;
d[i]=inf;
}
d[0]=0;
int num=0;
while(1){
int v=-1;
for(int i=1;i<=n;i++)
{
if(!vis[i]&&(v==-1||d[i]<d[v]))//找到最小的边
v=i;
}
if(v==-1)break;//无更新退出
vis[v]=1;
if(num&&cost[x[v]][y[v]])printf("%d %d\n",x[v],y[v]);
num=1;
for(int i=1;i<=n;i++)
{
if(d[i]>cost[i][v])//把所有的边都更新一边
{
d[i]=cost[i][v];
x[i]=i;
y[i]=v;
/* if(cost[i][v]!=0)
printf("%d %d\n",i,v);//每次更新全部时,即使不是下一个节点,也是会被更新的,所以要只输出下一个节点的*/
}
}
}
}
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%lf%lf",&e[i].x,&e[i].y);
for(int i=1;i<=n;i++)
{
for(int j=i+1;j<=n;j++)
{
cost[i][j]=cost[j][i]=dis(e[i],e[j]);
}
cost[i][i]=0;
}
scanf("%d",&m);
for(int i=1;i<=m;i++)
{
int a,b;
scanf("%d%d",&a,&b);
cost[a][b]=cost[b][a]=0;
}
prim();
return 0;
}
最新文章
- 学习web前端前感
- (高精度运算4.7.26)POJ 1220 NUMBER BASE CONVERSION(高精度数的任意进制的转换——方法:ba1----->;10进制----->;ba2)
- HW3.28
- C# 各版本更新简介
- 【C#】.NET中设置代理服务器浏览网页的实现--转载
- BZOJ 1631: [Usaco2007 Feb]Cow Party( 最短路 )
- HDU 4831 Scenic Popularity (段树)
- 依赖倒置原则DIP(面向接口编程—OOD)
- 修改NavigationBar样式
- URI和URL差别以及相对路径和绝对路径的差别
- jmeter中的函数
- Hbaseflush处理流程
- Python开发【模块】:tornado.queues协程的队列
- Google Protocol Buffers 入门
- VS与Opencv的亲密接触之安装配置过程
- Spring Cloud Dalston.SR5 BUG一记
- autolayout不work
- AE和Mocha结合做视频后期制作
- maven仓库介绍 牛人博客
- 【LeetCode】Validate Binary Search Tree 二叉查找树的推断