题目来源：POJ - 1751 ；

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

题目大意：给出N，代表了有n个村庄。然后是n行。每行两个数代表第i个村庄的坐标。然后是一个数k。代表了已经有k个村庄间有路。后面k行代表着已经有路的a b 村庄。现在想要将各村庄连通。求出距离最短的方案。输出。

思路：用了两个结构体。node用于读入各村庄的坐标。node1用于记录各个村庄间的距离。

#include<cstdio> #include<iostream> #include<cstring> #include<cmath> #include<algorithm> #define MAX 99999 using namespace std; int pre[MAX]; int n,m; struct node { int x,y; }p[1000];//用于读入村庄的坐标 struct node1 { int x,y; double distance;//存储村庄间的距离 }pi[751*751];//注意下表要开足够。 void init(int n) { for(int i=1;i<=n;i++) { pre[i]=i; } }//初始化pre[]数组 int find(int x) { if(pre[x]==x) return x; return pre[x]=find(pre[x]); }//找出根节点 bool cmp (node1 x,node1 y) { return x.distance<y.distance; } void mix(int x,int y) { int fx=find(x); int fy=find(y); if(fx!=fy) { pre[fy]=fx; } }//合并集合 int main() { while(scanf("%d",&n)!=EOF) { memset(pre,0,sizeof(pre));//要不要都行。 init(n);//初始化pre[]数组 for(int i=1;i<=n;i++) { scanf("%d%d",&p[i].x,&p[i].y); }//读入坐标 int k,u,v; scanf("%d",&k); for(int i=0;i<k;i++) { scanf("%d%d",&u,&v); mix(u,v);//先将已经有路的村庄连接起来。 } int s=0; for(int i=1;i<n;i++) { for(int j=i+1;j<=n;j++) { pi[s].x =i; pi[s].y =j; pi[s++].distance=sqrt((p[i].x-p[j].x)*(p[i].x-p[j].x)+(p[i].y-p[j].y)*(p[i].y-p[j].y)); }//用坐标计算出各村庄间的距离=根号下dx*dx+dy+dy } sort(pi,pi+s,cmp); for(int i=0;i<s;i++) { if(find(pi[i].x )!=find(pi[i].y )) { printf("%d %d\n",pi[i].x ,pi[i].y );//输出方案 mix(pi[i].x ,pi[i].y );//连接 } } } return 0; }

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