Highways「建议收藏」

Highways「建议收藏」HighwaysTimeLimit:1000MS MemoryLimit:10000KTotalSubmissions:14613 Accepted:4211 SpecialJudgeDescriptionTheislandnationofFlatopiaisperfectlyflat.

大家好,又见面了,我是你们的朋友全栈君。

Highways
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 14613   Accepted: 4211   Special Judge

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 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

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

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

code:

prim算法

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<sstream>
#include<algorithm>
#include<math.h>
#include<queue>
#include<stack>
#include<map>
#include<set>
using namespace std;
const int maxn=755;
const int INF=1000000;
double graph[maxn][maxn];
double lowcost[maxn];/*lowcost表示每个点的最小花费;*/
int closet[maxn];/*closet表示最小花费对应相连的点*/
int visited[maxn];/*visited区分两个集合*/
int n;/*n个点*/

struct dot{
    double x,y;
}a[maxn];

double f(dot p,dot q){
    return sqrt((p.x-q.x)*(p.x-q.x)+(p.y-q.y)*(p.y-q.y));
}

void createGraph(){
    memset(graph,0,sizeof(graph));
    memset(lowcost,0,sizeof(lowcost));
    memset(closet,0,sizeof(closet));
    memset(visited,0,sizeof(visited));
    for(int i=0;i<n;i++)
    for(int j=0;j<n;j++){
        if(i==j) graph[i][j]=INF;
        else graph[i][j]=graph[j][i]=f(a[i],a[j]);
    }
}

void prim(){
    visited[0]=1;/*选中第一个点*/
    for(int i=0;i<n;i++){
        lowcost[i]=graph[i][0];/*每个点与第一个点的权值*/
        closet[i]=0;/*与i点相连的是第一个点*/
    }
    for(int i=1;i<n;i++){  /*剩下n-1个点*/
        int k=0;
        double minn=lowcost[0];
        for(int j=0;j<n;j++){
            if(!visited[j] && lowcost[j]<minn){
                minn=lowcost[j];
                k=j;
            }
        }
        if(graph[k][closet[k]]!=0) printf("%d %d\n",k+1,closet[k]+1);
        visited[k]=1;
        for(int t=0;t<n;t++){  /*松弛操作*/
            if(!visited[t] && lowcost[t]>graph[t][k]){
                lowcost[t]=graph[t][k];
                closet[t]=k;
            }
        }
    }
}

int main()
{
 //   freopen("input.txt","r",stdin);
    int m;
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%lf%lf",&a[i].x,&a[i].y);
    }
    createGraph();
    scanf("%d",&m);
    int b,c;
    while(m--){
        scanf("%d%d",&b,&c);
        graph[b-1][c-1]=graph[c-1][b-1]=0;
    }
    prim();
}

Kruskal算法:注意要用G++提交

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<sstream>
#include<algorithm>
#include<math.h>
#include<queue>
#include<stack>
#include<map>
#include<set>
using namespace std;
const int MAXN = 755; /*结点数目上限*/
int pa[MAXN];    /*pa[x]表示x的父节点*/
int rank[MAXN];    /*rank[x]是x的高度的一个上界*/
int flag;
struct node{
    int x,y;
    double w;
}edge[MAXN*MAXN];

struct dot{
    double x,y;
}a[MAXN];

bool cmp(node p,node q){
    return p.w<q.w;
}

double f(dot p,dot q){
    return sqrt((p.x-q.x)*(p.x-q.x)+(p.y-q.y)*(p.y-q.y));
}

/*创建一个单元集*/
void make_set(int x)
{
    pa[x] = x;
    rank[x] = 0;
}

/*带路径压缩的查找*/
int find_set(int x)
{
    if(x != pa[x])
        pa[x] = find_set(pa[x]);
    return pa[x];
}

/*按秩合并x,y所在的集合*/
void union_set(int xx, int yy,double w)
{
    int x = find_set(xx);
    int y = find_set(yy);
    if(x == y)return ;
    if(rank[x] > rank[y])/*让rank比较高的作为父结点*/
    {
        pa[y] = x;
    }
    else
    {
        pa[x] = y;
        if(rank[x] == rank[y])
            rank[y]++;
    }
    if(w!=0){
        printf("%d %d\n",xx,yy);
        flag=1;
    }
}

int main()
{
   // freopen("input.txt","r",stdin);
    int n,m;
    scanf("%d",&n);
    for(int i=0;i<MAXN;i++)
        make_set(i);
    for(int i=1;i<=n;i++){
        scanf("%lf%lf",&a[i].x,&a[i].y);
    }
    int k=0;
    for(int i=1;i<=n;i++){
        for(int j=i+1;j<=n;j++){
            edge[k].x=i;
            edge[k].y=j;
            edge[k].w=f(a[i],a[j]);
            k++;
        }
    }
    scanf("%d",&m);
    int b,c;
    while(m--){
        scanf("%d%d",&b,&c);
        union_set(b,c,0);
    }
    sort(edge,edge+k,cmp);
    flag=0;
    for(int i=0;i<k;i++){
        union_set(edge[i].x,edge[i].y,edge[i].w);
    }
    if(!flag) printf("\n");
}

版权声明:本文内容由互联网用户自发贡献,该文观点仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 举报,一经查实,本站将立刻删除。

发布者:全栈程序员-站长,转载请注明出处:https://javaforall.net/129970.html原文链接:https://javaforall.net

赞 (0)
全栈程序员-站长的头像全栈程序员-站长

相关推荐

  • base64编码图片数据存储服务器

    base64编码图片数据存储服务器如果直接提交base64编码图片数据,过大的话后台会出现转发错误问题。我在刚开始接触base64编码图片数据时,就是把base64编码图片数据传到后台来解码生成图片。导致生成的图片无法打开,后来才发现其实传到后台的base64编码根本就不完整,导致解码出现问题,无法显示图片。所以,base64编码只能在前端处理。后来查阅资料,看见一个不错的解决方式就是

    2022年4月13日
  • react中类组件传值,函数组件传值:父子组件传值、非父子组件传值[通俗易懂]

    react中类组件传值,函数组件传值:父子组件传值、非父子组件传值[通俗易懂]react中父子组件传值

    2022年5月17日
  • 电子元器件常用品牌汇总(持续更新)

    电子元器件常用品牌汇总(持续更新)电阻:Yageo国巨、Fenghua风华、Rohm罗姆、TDK、Samsung三星、Uniohm厚声、Walsin华新科、Ralec旺诠、KOA兴亚、Panasonic松下、AVX、TMTEC泰铭、Kyocera京瓷、PHYCOM飞元。电容:Yageo国巨、Fenghua风华、Murata村田、TDK、Samsung三星、Eyang宇阳、Taiyo太诱、Kyocera京瓷、HEC禾伸堂、Kemet基美、ISND华信安、AVX、Nichicon尼吉康、Panasonic松下、SANYO三洋。电感:Mura

    2022年6月29日
  • rbac权限管理设计 7表_数据库角色权限表设计

    rbac权限管理设计 7表_数据库角色权限表设计RBAC(Role-BasedAccessControl,基于角色的访问控制),就是用户通过角色与权限进行关联。简单地说,一个用户拥有若干角色,每一个角色拥有若干权限。这样,就构造成“用户-角色-权限”的授权模型。在这种模型中,用户与角色之间,角色与权限之间,一般者是多对多的关系。(如下图)当用户量非常多的时候,逐一的给用户授权角色是一件很痛苦的事情,于是引出组的概念。

    2022年9月29日
  • 【Maven】maven打包生成可执行jar文件

    【Maven】maven打包生成可执行jar文件maven默认打包生成的jar是不能够直接运行的,因为在jar文件的META-INF/MANIFEST.MF文中没有Main-Class一行,为了生成可执行的jar文件,需要借助maven的插件,maven-shade-plugin,配置该插件如下:

    2022年5月25日
  • CRC校验算法详解「建议收藏」

    CRC校验算法详解「建议收藏」CRC(CyclicRedundancyCheck)循环冗余校验是常用的数据校验方法,讲CRC算法的文章很多,之所以还要写这篇,是想换一个方法介绍CRC算法,希望能让大家更容易理解CRC算法。先

    2022年8月2日

发表回复

您的邮箱地址不会被公开。 必填项已用 * 标注

关注全栈程序员社区公众号