zoukankan      html  css  js  c++  java
  • POJ3625 Building Roads

     
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 10803   Accepted: 3062

    Description

    Farmer John had just acquired several new farms! He wants to connect the farms with roads so that he can travel from any farm to any other farm via a sequence of roads; roads already connect some of the farms.

    Each of the N (1 ≤ N ≤ 1,000) farms (conveniently numbered 1..N) is represented by a position (XiYi) on the plane (0 ≤ X≤ 1,000,000; 0 ≤ Y≤ 1,000,000). Given the preexisting M roads (1 ≤ M ≤ 1,000) as pairs of connected farms, help Farmer John determine the smallest length of additional roads he must build to connect all his farms.

    Input

    * Line 1: Two space-separated integers: N and M
    * Lines 2..N+1: Two space-separated integers: Xand Y
    * Lines N+2..N+M+2: Two space-separated integers: i and j, indicating that there is already a road connecting the farm i and farm j.

    Output

    * Line 1: Smallest length of additional roads required to connect all farms, printed without rounding to two decimal places. Be sure to calculate distances as 64-bit floating point numbers.

    Sample Input

    4 1
    1 1
    3 1
    2 3
    4 3
    1 4

    Sample Output

    4.00

    Source

     
    kruskal最小生成树
    先处理出每两个点之间的距离,加入到边集。读入的已连接的点对也加入边集,距离为0。跑一遍kruskal出解
     
     1 /*by SilverN*/
     2 #include<algorithm>
     3 #include<iostream>
     4 #include<cstring>
     5 #include<cstdio>
     6 #include<cmath>
     7 using namespace std;
     8 const int mxn=1020;
     9 int n,m;
    10 struct point{
    11     double x,y;
    12 }p[mxn];
    13 struct edge{
    14     int x,y;
    15     double dis;
    16 }e[mxn*mxn];
    17 int cmp(edge a,edge b){
    18     return a.dis<b.dis;
    19 }
    20 double dist(point a,point b){
    21     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
    22 }
    23 //
    24 int fa[mxn];
    25 int find(int x){
    26     if(fa[x]==x)return x;
    27     return fa[x]=find(fa[x]);
    28 }
    29 int cnt=0;
    30 double ans=0;
    31 void kruskal(){
    32     int now=0;
    33     for(int i=1;i<=cnt && now<n-1;i++){
    34         int x=find(e[i].x);
    35         int y=find(e[i].y);
    36         if(x!=y){
    37             ans+=e[i].dis;
    38             fa[x]=y;now++;
    39         }
    40     }
    41     printf("%.2f
    ",ans);
    42     return;
    43 }
    44 int main(){
    45     scanf("%d%d",&n,&m);
    46     int i,j;
    47     for(i=1;i<=n;i++) fa[i]=i;
    48     for(i=1;i<=n;i++)
    49         scanf("%lf%lf",&p[i].x,&p[i].y);
    50     for(i=1;i<n;i++)
    51       for(j=i+1;j<=n;j++){
    52           e[++cnt].dis=dist(p[i],p[j]);
    53           e[cnt].x=i;e[cnt].y=j;
    54       }
    55     for(i=1;i<=m;i++){
    56         cnt++;
    57         scanf("%d%d",&e[cnt].x,&e[cnt].y);
    58         e[cnt].dis=0;
    59     }
    60     sort(e+1,e+cnt+1,cmp);
    61     kruskal();
    62     return 0;
    63 }
  • 相关阅读:
    使用 Fetch
    实现一个联系客服对话框的前端部分
    javascript之Object.defineProperty的奥妙
    vue之nextTick全面解析
    创建元素和删除元素
    vue.js应用开发笔记
    待字闺中之最多连续数的子集
    HDU-1212-Big Number
    虚方法【仅仅有虚方法或者抽象方法才干被子类方法重写】
    利用localStorage实现对ueditor编辑内容定时保存为草稿
  • 原文地址:https://www.cnblogs.com/SilverNebula/p/5872558.html
Copyright © 2011-2022 走看看