Agri-Net_usaco3.1.1_poj1258_最小生成树

Description

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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

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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

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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.

Source

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USACO 102

题目大意

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给出n*n的邻接矩阵，求这n个点的最小生成树

思路

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因为我比较懒所以最小生成树用了kruskal，写了邻接表+并查集优化
不是很理解为什么写c++比较优美了现在

代码/pas

–

type
  edge=record
    x,y,w:longint;
  end;
var
  a,b,n,i,k:longint;
  f,v:array[1..200]of longint;
  e:array[1..10000]of edge;
  min,ans:longint;
procedure qsort(l,r:longint);
var
  x,y,key:longint;
  temp:edge;
begin
  if l>=r then exit;
  x:=l;
  y:=r;
  key:=e[l+random(r-l+1)].w;
  repeat
    while  (e[x].w<key) do inc(x);
    while  (e[y].w>key) do dec(y);
    if x<=y then
    begin
      temp:=e[x];
      e[x]:=e[y];
      e[y]:=temp;
      inc(x);
      dec(y);
    end;
  until x>y;
  qsort(l,y);
  qsort(x,r);
end;
function find(x:longint):longint;
var
  y,w,root:longint;
begin
  y:=x;
  while f[y]<>0 do y:=f[y];
  root:=y;
  y:=x;
  while f[y]<>0 do
  begin
    w:=f[y];
    f[y]:=root;
    y:=w;
  end;
  find:=root;
end;
procedure union(x,y:longint);
begin
  if find(x)<>find(y) then
  f[x]:=y;
end;
procedure init;
var
  i,j:Longint;
begin
  readln(n);
  for i:=1 to n do
  for j:=1 to n do
  with e[(i-1)*n+j] do
  begin
    read(w);
    x:=i;
    y:=j;
  end;
  for i:=1 to n do v[i]:=i;
  fillchar(f,sizeof(f),0);
  ans:=0;
end;
begin
  while not seekeof do
  begin
    init;
    qsort(1,n*n);
    k:=0; 
    for i:=1 to n*n-1 do
    if k=n then break
    else
    begin
      a:=find(e[i].x);
      b:=find(e[i].y);
      if a<>b then
      begin
        ans:=ans+e[i].w;
        union(a,b);
        inc(k);
      end;
    end;
    writeln(ans);
  end;
end.

代码/c++:

/*
ID:wjp13241
PROG:agrinet
LANG:C++
*/
#include <stdio.h>
#include <algorithm>
using namespace std;
struct edge
{
    int x,y,w;
};
edge e[10002];
int f[101];
int maxE=0;
bool cmp(edge x,edge y)
{
    return x.w<y.w;
}
void add(int x,int y,int w)
{
    e[++maxE]=(edge){x,y,w};
}
int find(int x)
{
    if (!f[x])
        return x;
    f[x]=find(f[x]);
    return f[x];
}
int main()
{
    freopen("agrinet.in","r",stdin);
    freopen("agrinet.out","w",stdout);
    int n,k=0,ans=0;
    scanf("%d",&n);
    for (int i=1;i<=n;i++)
        for (int j=1;j<=n;j++)
        {
            int tmp;
            scanf("%d",&tmp);
            if (tmp)
                add(i,j,tmp);
        }
    sort(e+1,e+maxE+1,cmp);
    for (int i=1;i<maxE;i++)
    {
        if (find(e[i].x)!=find(e[i].y))
        {
            f[find(e[i].x)]=find(e[i].y);
            ans+=e[i].w;
            k++;
        }
        if (k==n-1)
            break;
    }
    printf("%d\n",ans);
    return 0;
}