Highways
Time Limit: 1000MSMemory Limit: 10000K
Total Submissions: 9654Accepted: 2725Special 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 ith 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
91 50 0 3 24 55 10 45 21 25 331 39 71 2
Sample Output
1 63 74 95 78 3
Source
题目链接：?id=1751题目大意：有n个村庄，给出它们的坐标，已经有m条路，连接两个村庄，问要使得任意两个村庄连通并且路的总距离最短，如何建路，输出路径题目分析：建个图，将已经连通的点合并，，然后跑一下Kruskal记录一下路径即可#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int const MAX = 800;int fa[MAX], re1[MAX], re2[MAX];int n, m, cnt, num;struct Node{int u, v;}nd[MAX];struct Edge{int u, v, w;}e[MAX * MAX / 2];bool cmp(Edge a, Edge b){return a.w < b.w;}void UF_set(){for(int i = 0; i <= MAX; i++)fa[i] = i;}int Find(int x){return x == fa[x] ? x : fa[x] = Find(fa[x]);}void Union(int a, int b){int r1 = Find(a);int r2 = Find(b);if(r1 != r2)fa[r2] = r1;}int dis(int x1, int y1, int x2, int y2){return (x1 – x2) * (x1 – x2) + (y1 – y2) * (y1 – y2);}void Kruskal(){num = 0;for(int i = 0; i < cnt; i++){int u = e[i].u;int v = e[i].v;if(Find(u) != Find(v)){Union(u, v);re1[num] = u + 1;re2[num] = v + 1;num ++;}if(num >= n – 1)break;}}int main(){cnt = 0;scanf("%d", &n);for(int i = 0; i < n; i++)scanf("%d %d", &nd[i].u, &nd[i].v);for(int i = 0; i < n; i++){for(int j = i + 1; j < n; j++){e[cnt].u = i;e[cnt].v = j;e[cnt++].w = dis(nd[i].u, nd[i].v, nd[j].u, nd[j].v);}}sort(e, e + cnt, cmp);scanf("%d", &m);UF_set();for(int i = 0; i < m; i++){int u, v;scanf("%d %d", &u, &v);u –;v –;if(Find(u) != Find(v))Union(u, v);}Kruskal();for(int i = 0; i < num; i++)printf("%d %d\n", re1[i], re2[i]);}
接受失败也等于给了自己从零开始的机会，接受失败更是一种智者的宣言和呐喊;
