POJ1751 Highways(Prim)

Highways

Description

The island nation of Flatopia is perfectly flat. Unfortunately, Flatopia has a very poor system of public highways. The Flatopian government is aware of this problem and has already constructed a number of highways connecting some of the most important towns. However, there are still some towns that you can’t reach via a highway. It is necessary to build more highways so that it will be possible to drive between any pair of towns without leaving the highway system.

Flatopian towns are numbered from 1 to N and town i has a position
given by the Cartesian coordinates (xi, yi). Each highway connects
exaclty two towns. All highways (both the original ones and the ones
that are to be built) follow straight lines, and thus their length is
equal to Cartesian distance between towns. All highways can be used in
both directions. Highways can freely cross each other, but a driver can
only switch between highways at a town that is located at the end of
both highways.

The Flatopian government wants to minimize the cost of building new
highways. However, they want to guarantee that every town is
highway-reachable from every other town. Since Flatopia is so flat, the
cost of a highway is always proportional to its length. Thus, the least
expensive highway system will be the one that minimizes the total
highways length.

Input

The
input consists of two parts. The first part describes all towns in the
country, and the second part describes all of the highways that have
already been built.

The first line of the input file contains a single integer N (1
<= N <= 750), representing the number of towns. The next N lines
each contain two integers, xi and yi separated by a space. These values
give the coordinates of i^th town (for i from 1 to N). Coordinates will have an absolute value no greater than 10000. Every town has a unique location.

The next line contains a single integer M (0 <= M <= 1000),
representing the number of existing highways. The next M lines each
contain a pair of integers separated by a space. These two integers give
a pair of town numbers which are already connected by a highway. Each
pair of towns is connected by at most one highway.

Output

Write
to the output a single line for each new highway that should be built in
order to connect all towns with minimal possible total length of new
highways. Each highway should be presented by printing town numbers that
this highway connects, separated by a space.

If no new highways need to be built (all towns are already
connected), then the output file should be created but it should be
empty.

Sample Input

9
1 5
0 0
3 2
4 5
5 1
0 4
5 2
1 2
5 3
3
1 3
9 7
1 2

Sample Output

1 6
3 7
4 9
5 7
8 3
【题意】给你N个城市的坐标，城市之间存在公路，但是由于其中一些道路损坏了，需要维修，维修的费用与公路长成正比（公路是直的）。
但现有M条公路是完整的，不需要维修，下面有M行，表示不需要维修的道路两端的城市，问最短费用。
【分析】一道典型的最小生成树题，lowcost[i]数组存还未处理的城市i离已经处理过的城市的最短距离,pre[i]]数组存还未处理的
城市i离已经处理过的哪个城市最近。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#include<functional>
#define mod 1000000007
#define inf 0x3f3f3f3f
#define pi acos(-1.0)
using namespace std;
typedef long long ll;
const int N=;
const int M=;
int n,m,k,edg[N][N],x[N],y[N],lowcost[N],pre[N];
void Prim() {
  for(int i=;i<=n;i++){
    lowcost[i]=edg[][i];
    pre[i]=;
  }
  lowcost[]=-;
  for(int i=;i<n;i++){
      int minn=inf;
      for(int j=;j<=n;j++){
        if(lowcost[j]!=-&&lowcost[j]<minn){
            minn=lowcost[j];
            k=j;
        }
      }
      if(lowcost[k]!=)printf("%d %d\n",pre[k],k);
      lowcost[k]=-;
      for(int j=;j<=n;j++){
        if(edg[j][k]<lowcost[j]){
            lowcost[j]=edg[j][k];
            pre[j]=k;
        }
      }
  }
}
int main() {
    while(~scanf("%d",&n)) {
        for(int i=; i<=n; i++) {
            scanf("%d%d",&x[i],&y[i]);
            for(int j=; j<i; j++)
               edg[i][j]=edg[j][i]=(x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]);
            edg[i][i]=inf;
        }
        scanf("%d",&m);
        for(int i=; i<m; i++) {
            int ita,itb;
            scanf("%d%d",&ita,&itb);
            edg[ita][itb]=edg[itb][ita]=;
        }
        Prim();
    }
    return ;
}