zoukankan      html  css  js  c++  java
  • poj 1258 Agri-Net

    Agri-Net
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 44147   Accepted: 18054

    Description

    Farmer John has been elected mayor of his town! One of his campaign promises was to bring internet connectivity to all farms in the area. He needs your help, of course. 
    Farmer John ordered a high speed connection for his farm and is going to share his connectivity with the other farmers. To minimize cost, he wants to lay the minimum amount of optical fiber to connect his farm to all the other farms. 
    Given a list of how much fiber it takes to connect each pair of farms, you must find the minimum amount of fiber needed to connect them all together. Each farm must connect to some other farm such that a packet can flow from any one farm to any other farm. 
    The distance between any two farms will not exceed 100,000. 

    Input

    The input includes several cases. For each case, the first line contains the number of farms, N (3 <= N <= 100). The following lines contain the N x N conectivity matrix, where each element shows the distance from on farm to another. Logically, they are N lines of N space-separated integers. Physically, they are limited in length to 80 characters, so some lines continue onto others. Of course, the diagonal will be 0, since the distance from farm i to itself is not interesting for this problem.

    Output

    For each case, output a single integer length that is the sum of the minimum length of fiber required to connect the entire set of farms.

    Sample Input

    4
    0 4 9 21
    4 0 8 17
    9 8 0 16
    21 17 16 0
    

    Sample Output

    28

    Source

    prime做法:

     1 #include <cstdio>
     2 #include <cstring>
     3 #include <string>
     4 #include <queue>
     5 #include <stack>
     6 #include <iostream>
     7 using namespace std;
     8 #define lengthmax  100005
     9 int map[105][105];
    10 int dis[105];
    11 int main(){
    12     //freopen("D:\INPUT.txt","r",stdin);
    13     int n;
    14     while(scanf("%d",&n)!=EOF){
    15         int i,j;
    16         for(i=0;i<n;i++){
    17             dis[i]=lengthmax;
    18             for(j=0;j<n;j++){
    19                 scanf("%d",&map[i][j]);
    20             }
    21         }
    22         int m=1,mink,s=0,min,sum=0;
    23         dis[0]=0;//已经在集合里面
    24         while(m<n){//循环n-1次
    25             min=lengthmax;
    26             for(i=1;i<n;i++){
    27                 if(dis[i]>map[s][i]){//更新
    28                     dis[i]=map[s][i];
    29                 }
    30                 if(dis[i]&&min>dis[i]){
    31                     min=dis[i];
    32                     mink=i;
    33                 }
    34             }
    35             //cout<<mink<<endl;
    36             //cout<<min<<endl;
    37             sum+=min;
    38             dis[mink]=0;
    39             s=mink;
    40             m++;
    41         }
    42         cout<<sum<<endl;
    43     }
    44     return 0;
    45 }

    kruskal做法:

     1 #include <cstdio>
     2 #include <cstring>
     3 #include <string>
     4 #include <queue>
     5 #include <stack>
     6 #include <algorithm>
     7 #include <iostream>
     8 using namespace std;
     9 struct node{
    10     int u,v,l;
    11 };
    12 node p[5000];
    13 int fa[105];
    14 bool cmp(node a,node b){
    15     return a.l<b.l;
    16 }
    17 int findfa(int a){
    18     if(a!=fa[a]){
    19         fa[a]=findfa(fa[a]);//attention!
    20     }
    21     return fa[a];
    22 }
    23 int main(){
    24     //freopen("D:\INPUT.txt","r",stdin);
    25     int n;
    26     while(scanf("%d",&n)!=EOF){
    27         int i,j,k;
    28         for(i=0;i<n;i++){
    29             fa[i]=i;
    30         }
    31         int num;
    32         k=0;
    33         for(i=0;i<n;i++){
    34             for(j=0;j<n;j++){
    35                 scanf("%d",&num);
    36                 if(i<j){
    37                     p[k].u=i;
    38                     p[k].v=j;
    39                     p[k++].l=num;
    40                 }
    41             }
    42         }
    43         int sum=0;
    44         num=1;
    45         sort(p,p+k,cmp);
    46 
    47         for(i=0;i<k;i++){
    48             if(num==n){
    49                 break;
    50             }
    51             int u=findfa(p[i].u);
    52             int v=findfa(p[i].v);
    53             if(u==v){
    54                 continue;
    55             }
    56             if(u>v){
    57                 fa[v]=u;
    58             }
    59             else{
    60                 fa[u]=v;
    61             }
    62             sum+=p[i].l;
    63             num++;
    64         }
    65         cout<<sum<<endl;
    66     }
    67     return 0;
    68 }
  • 相关阅读:
    PHP使用引用变量foreach时,切记其他循环不要使用同一个名字的变量
    PHP 获取给定时间的周日时间或月末时间或每天
    MySQL Load Data InFile 文件内容导入数据库和 Into OutFile导出数据到文件
    直接拿来用!最火的iOS开源项目(一)
    12个有趣的C语言问答
    Flex,Flash,AS3,AIR的关系和区别
    Stage3D大冒险
    c/c++程序中内存区划分
    IOS—— strong weak retain assign 学习
    如何提高你的移动开发中AS3/AIR性能
  • 原文地址:https://www.cnblogs.com/Deribs4/p/4645114.html
Copyright © 2011-2022 走看看