zoukankan      html  css  js  c++  java
  • POJ 3265 Building Roads(最小生成树)

    Building Roads
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 9255   Accepted: 2669

    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

    用Prime算法会比较方便
     1 #include <iostream>
     2 #include <cstring>
     3 #include <cstdio>
     4 #include <string>
     5 #include <cmath>
     6 #include <iomanip>
     7 using namespace std;
     8 #define maxn 1010
     9 double mp[maxn][maxn];
    10 int N, M;
    11 int vis[maxn];
    12 double low[maxn];
    13 struct Point{
    14     double x,y;
    15 }p[maxn];
    16 #define INF 99999999
    17 double cal(double x1, double y1, double x2, double y2){
    18     return ( sqrt( (x1-x2)*(x1-x2) +(y1-y2)*(y1-y2)));
    19 }
    20 double prime(){
    21     double result;int pos;
    22     memset(vis, 0, sizeof(vis));
    23     pos = 1; vis[1] = 1; low[1] = 0;
    24     for(int i = 1; i <= N; i++){
    25         if(i != pos ) low[i] = mp[1][i];  
    26     }
    27     for(int i = 1; i <= N-1; i++){
    28         double min = INF;
    29         for(int j = 1; j <= N; j++){
    30             if(low[j] < min && !vis[j]){
    31                 min = low[j];
    32                 pos = j;
    33             }
    34         }
    35         result += min;
    36         vis[pos] = 1;
    37         
    38         for(int j = 1; j <= N; j++){
    39             if(!vis[j] && mp[pos][j] < low[j]) low[j] = mp[pos][j];
    40         }
    41     }
    42     return result;
    43     
    44 }
    45 int main(){
    46     while(~scanf("%d%d", &N, &M)){
    47         memset(mp, 0, sizeof(mp));
    48         for(int i = 1; i <= N; i++){
    49             scanf("%lf%lf", &p[i].x, &p[i].y);
    50         }
    51         for(int i = 1; i <= N; i++){
    52             for(int j = 1; j <= N; j++){
    53                 if(i != j){
    54                     mp[i][j] = mp[j][i] = cal(p[i].x, p[i].y,p[j].x, p[j].y);
    55                 }
    56             }
    57         }
    58         for(int i = 1; i <= M; i++){
    59             int t1, t2;
    60             scanf("%d%d", &t1, &t2);
    61             mp[t1][t2] = mp[t2][t1] = 0;
    62         }
    63         cout<<fixed<<setprecision(2)<<prime()<<endl;
    64     //    printf("%lf
    ", prime());
    65     }
    66     
    67     return 0;
    68 }


  • 相关阅读:
    PTA 7-29 修理牧场(Huffman树)
    全网最详细最好懂 PyTorch CNN案例分析 识别手写数字
    【Python Deap库】遗传算法/遗传编程 进化算法基于python DEAP库深度解析讲解
    【比较】遗传算法GA和遗传编程GP有什么不同?
    【python(deap库)实现】GEAP 遗传算法/遗传编程 genetic programming +
    【比较】粒子群算法PSO 和 遗传算法GA 的相同点和不同点
    【遗传编程/基因规划】Genetic Programming
    【经典大数据竞赛科普】泰坦尼克灾难 到底是个什么东西
    【Python代码】TSNE高维数据降维可视化工具 + python实现
    【python代码】 最大流问题+最小花费问题+python(ortool库)实现
  • 原文地址:https://www.cnblogs.com/titicia/p/3917779.html
Copyright © 2011-2022 走看看