HDU 1162 Eddy & #39; s picture (graph theory-Minimum Spanning Tree), hdueddy

HDU 1162 Eddy's picture (graph theory-Minimum Spanning Tree), hdueddy

The questions are as follows:

Eddy's picture Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission (s): 7267 Accepted Submission (s): 3676


Problem DescriptionEddy begins to like painting pictures recently, he is sure of himself to become a painter. every day Eddy draws pictures in his small room, and he usually puts out his newest pictures to let his friends appreciate. but the result it can be imagined, the friends are not interested in his picture. eddy feels very puzzled, in order to change all friends's view to his technical of painting pictures, so Eddy creates a problem for the his friends of you.
Problem descriptions as follows: Given you some coordinates pionts on a drawing paper, every point links with the ink with the straight line, causes all points finally to link in the same place. how many distants does your duty discover the shortest length which the ink draws?
 
InputThe first line contains 0 <n <= 100, the number of point. for each point, a line follows; each following line contains two real numbers indicating the (x, y) coordinates of the point.

Input contains multiple test cases. Process to the end of file.
 
OutputYour program prints a single real number to two decimal places: the minimum total length of ink lines that can connect all the points.
 
Sample Input

31.0 1.02.0 2.02.0 4.0

 
Sample Output

3.41

 

Question:

Given the coordinates of n points (two-dimensional x, y), you can use a straight line (undirected) to connect any two points, connect all points to form a whole (so that any two points can be reached), the minimum distance of the line.

A simple Minimum Spanning Tree problem, the right here should be obtained based on the coordinate information of the Input Point. When the input is stored, the subscript in the array of the point can be considered as the number of the point, we can use the complexity of O (n ^ 2) to calculate the distance between any two different points as the weight. The two vertices of the edge are the numbers of the points at both ends of the distance. The next step is the simple Minimum Spanning Tree Algorithm. You can use kruskal or prim.

Implementation of the kruskal Algorithm

#include <cstdio>#include <vector>#include <algorithm>#include <cmath>using namespace std;const int N = 105;struct Edge{int x, y;double w;};struct Point{double x;double y;};int pre[N];Point point[N];Edge edges[N * N / 2];int i_p, i_e, cnt;double res;int root(int x){if (x != pre[x]){pre[x] = root(pre[x]);}return pre[x];}bool merge(int x, int y){int fx = root(x);int fy = root(y);bool ret = false;if (fx != fy){pre[fx] = pre[fy];ret = true;--cnt;}return ret;}void init(int n){cnt = n;res = 0;for (int i = 0; i <= n; ++i){pre[i] = i;}}bool cmp(const Edge &a, const Edge &b){return a.w < b.w;}int main(){int n;double dx, dy;while (scanf("%d", &n) != EOF){init(n);i_e = i_p = 0;for (int i = 0; i < n; ++i){scanf("%lf %lf", &dx, &dy);point[i_p].x = dx;point[i_p].y = dy;++i_p;}for (int i = 0; i < n; ++i){for (int j = i + 1; j < n; ++j){edges[i_e].x = i;edges[i_e].y = j;double dd = (point[i].x - point[j].x) * (point[i].x - point[j].x);dd += (point[i].y - point[j].y) * (point[i].y - point[j].y);edges[i_e].w = sqrt(dd);++i_e;}}sort(edges, edges + i_e, cmp);//the cnt == 1 indicates that the mixnum spanning tree is builded sucessfully.for (int i = 0; i < i_e && cnt != 1; ++i){if (merge(edges[i].x, edges[i].y))res += edges[i].w;}printf("%.2lf\n", res);}return 0;}

Prim

#include <cstdio>#include <cmath>#include <string>#include <climits>using namespace std;const int N = 105;struct Point{double x;double y;};Point point[N];double dist[N], graph[N][N];bool visit[N];int i_p, i_e;double res;double prim(int n){for (int i = 0; i < n; ++i){dist[i] = graph[1][i];}visit[1] = true;res = 0;for (int i = 1; i < n; ++i){int min_label;double minx = INT_MAX * 1.0;for (int j = 0; j < n; ++j){if (visit[j])continue;if (minx > dist[j]){minx = dist[j];min_label = j;}}visit[min_label] = true;res += minx;for (int j = 0; j < n; ++j){if (!visit[j]){if (dist[j] > graph[min_label][j]){dist[j] = graph[min_label][j];}}}}return res;}void init(int n){memset(visit, 0, sizeof(visit));}int main(){int n;double dx, dy;while (scanf("%d", &n) != EOF){i_e = i_p = 0;res = 0;memset(visit, 0, sizeof(visit));for (int i = 0; i < n; ++i){scanf("%lf %lf", &dx, &dy);point[i_p].x = dx;point[i_p].y = dy;++i_p;}for (int i = 0; i < n; ++i){for (int j = i + 1; j < n; ++j){double dd = (point[i].x - point[j].x) * (point[i].x - point[j].x);dd += (point[i].y - point[j].y) * (point[i].y - point[j].y);graph[i][j] = graph[j][i] = sqrt(dd);}}res = prim(n);printf("%.2lf\n", res);}return 0;}