POJ1751 Highways(Prim)

时间：2016-08-03 10:34:06 阅读：118 评论：0 收藏：0 [点我收藏+]

标签：

Highways

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 13182		Accepted: 3814		Special Judge

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

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=10005;
const int M=15005;
int n,m,k,edg[N][N],x[N],y[N],lowcost[N],pre[N];
void Prim() {
  for(int i=1;i<=n;i++){
    lowcost[i]=edg[1][i];
    pre[i]=1;
  }
  lowcost[1]=-1;
  for(int i=1;i<n;i++){
      int minn=inf;
      for(int j=1;j<=n;j++){
        if(lowcost[j]!=-1&&lowcost[j]<minn){
            minn=lowcost[j];
            k=j;
        }
      }
      if(lowcost[k]!=0)printf("%d %d\n",pre[k],k);
      lowcost[k]=-1;
      for(int j=1;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=1; i<=n; i++) {
            scanf("%d%d",&x[i],&y[i]);
            for(int j=1; 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=0; i<m; i++) {
            int ita,itb;
            scanf("%d%d",&ita,&itb);
            edg[ita][itb]=edg[itb][ita]=0;
        }
        Prim();
    }
    return 0;
}

View Code

POJ1751 Highways(Prim)

标签：

原文地址：http://www.cnblogs.com/jianrenfang/p/5731698.html

踩

(0)

评论一句话评论（0）

分享档案

更多>

2021年07月29日 (22)
2021年07月28日 (40)
2021年07月27日 (32)
2021年07月26日 (79)
2021年07月23日 (29)
2021年07月22日 (30)
2021年07月21日 (42)
2021年07月20日 (16)
2021年07月19日 (90)
2021年07月16日 (35)

周排行