zoukankan      html  css  js  c++  java
  • POJ 1287 Networking

    Networking
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 4457   Accepted: 2421

    Description

    You are assigned to design network connections between certain points in a wide area. You are given a set of points in the area, and a set of possible routes for the cables that may connect pairs of points. For each possible route between two points, you are given the length of the cable that is needed to connect the points over that route. Note that there may exist many possible routes between two given points. It is assumed that the given possible routes connect (directly or indirectly) each two points in the area.
    Your task is to design the network for the area, so that there is a connection (direct or indirect) between every two points (i.e., all the points are interconnected, but not necessarily by a direct cable), and that the total length of the used cable is minimal.

    Input

    The input file consists of a number of data sets. Each data set defines one required network. The first line of the set contains two integers: the first defines the number P of the given points, and the second the number R of given routes between the points. The following R lines define the given routes between the points, each giving three integer numbers: the first two numbers identify the points, and the third gives the length of the route. The numbers are separated with white spaces. A data set giving only one number P=0 denotes the end of the input. The data sets are separated with an empty line.
    The maximal number of points is 50. The maximal length of a given route is 100. The number of possible routes is unlimited. The nodes are identified with integers between 1 and P (inclusive). The routes between two points i and j may be given as i j or as j i.

    Output

    For each data set, print one number on a separate line that gives the total length of the cable used for the entire designed network.

    Sample Input

    1 0
    
    2 3
    1 2 37
    2 1 17
    1 2 68
    
    3 7
    1 2 19
    2 3 11
    3 1 7
    1 3 5
    2 3 89
    3 1 91
    1 2 32
    
    5 7
    1 2 5
    2 3 7
    2 4 8
    4 5 11
    3 5 10
    1 5 6
    4 2 12
    
    0

    Sample Output

    0
    17
    16
    26

    Source

    //最小生成树模板题,呵呵

    #include
    <iostream> #include <stdio.h> #include <string.h> #include <algorithm> #include <cmath> using namespace std; int g[53][53]; bool b[53]; int p,n; void prm() { memset(b,0,sizeof(b)); int i,j,min,t=p; int sum=0; while(--t) { min=100000; for(i=2;i<=p;i++) if(!b[i]&&min>g[1][i]) { j=i; min=g[1][j]; } sum+=min; b[j]=1; for(i=2;i<=p;i++) if(!b[i]&&g[1][i]>g[j][i]) g[1][i]=g[j][i]; } printf("%d\n",sum); } int main() { int u,v,cost; while(scanf("%d%d",&p,&n),p) { for(u=1;u<=p;u++) for(v=1;v<=p;v++) g[u][v]=1000; while(n--) { scanf("%d%d%d",&u,&v,&cost); g[u][v]=g[u][v]>cost?cost:g[u][v]; g[v][u]=g[u][v]; } prm(); } return 0; }
  • 相关阅读:
    策略模式
    Java反射机制
    两个无符号的正大数相加
    MySQL大表优化方案
    造成mysql慢查询的原因
    mysql对于很长的字符列的索引方案
    lyt经典版MySQL基础——函数
    lyt经典版MySQL基础——存储过程
    lyt经典版MySQL基础——变量
    lyt经典版MySQL基础——视图
  • 原文地址:https://www.cnblogs.com/372465774y/p/2592557.html
Copyright © 2011-2022 走看看