最小生成树——Highways（普利姆算法结合对于父亲节点标记）

题意描述：
高速路修路问题，有一些点是可以直接连通的现在需要修其他的路径不过现在要的不是总的路径长度，而是要修的每一条路的顶点编号
原题：
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

解题思路：
重新定义一个数组p来表示更新节点的父亲节点例如：
这样一组数据：
1 2 1
1 3 2
2 4 6
3 4 2
在最开始的时候都把父亲节点定为1，（下面就是普利姆算法）然后找距离1最近的点然后找到了节点2然后就该输出这个点和他的父亲节点;-----原因：（具体分析看代码中的讲述）
先看这一部分代码：

            min=inf;for(i=1; i<=n; i++){if(book[i]==0&&dis[i]<min){min=dis[i]; j=i;}}book[j]=1;//在这个地方找到的已经是当前距离根节点最近的点也就是说这条路是一定要修的//可以想一下，以前我们求距离的时候在这个地方把他们的距离加上了这一次我们//同样根据题目要求的不同吧要求的距离换成了两节点的信息（应该可以懂了吧）if(e[p[j]][j]!=0)printf("%d %d\n", p[j],j);

开始寻找2的出边找到了4，当时book[4]==0又因为dis[4]当时的数值是无穷大所以就更新为6，也就是下面的操作

             for(k=1; k<=n; k++){if(book[k]==0&&dis[k]>e[j][k]){dis[k]=e[j][k];p[k]=j;//找到要更新的节点是k节点//上一步找到距离源节点最近的点是j//然后再通过j节点并且作比较的也是e[j][k](代表j到//k的距离也就是说j和k是连接的）来更新dis[k]的，//现在应该理解当前k的父亲节点就是j}}

每次对于更新的点不能保证更新的最近的节点是dis[k],所以我们是在上一步直到j节点更新为最近节点的时候才把它和它的父亲节点输出（说实话做这道题的时候想了很具就是想不明白父亲节点的更换甚至当时还想到用结构体来记录父亲节点但是显然失败了因为出现的距离可能相等）
下面是AC代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
int e[2000][2000];
int dis[20000],p[2000];
int book[20000];
int inf=99999999;
struct node{int x;int y;
}s[2000];
int main()
{int n,m,t,min,i,j,k,a,b,w,r;while(scanf("%d", &n)!=EOF){for(i=1; i<=n; i++)scanf("%d %d", &s[i].x,&s[i].y);for(i=1; i<=n; i++){for(j=1; j<=n; j++){w=s[i].x-s[j].x;r=s[i].y-s[j].y;e[i][j]=e[j][i]=w*w+r*r;}} scanf("%d", &m);for(i=1; i<=m; i++){scanf("%d %d", &a,&b);e[a][b]=e[b][a]=0;}for(i=1; i<=n; i++)p[i]=1;for(i=1; i<=n; i++)dis[i]=e[1][i];memset(book,0,sizeof(book));book[1]=1;for(t=1; t<n; t++){min=inf;for(i=1; i<=n; i++){if(book[i]==0&&dis[i]<min){min=dis[i]; j=i;}}book[j]=1;if(e[p[j]][j]!=0)printf("%d %d\n", p[j],j);for(k=1; k<=n; k++){if(book[k]==0&&dis[k]>e[j][k]){dis[k]=e[j][k];p[k]=j;}}}}return 0;}

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最小生成树——Highways（普利姆算法结合对于父亲节点标记）

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