题目链接
题目分析
最小生成树问题;
解题思路
暴力计算任两点的距离(边权);
用Kruskal
算法 + 并查集 解决;
AC程序(C++)
/**********************************
*@ID: 3stone
*@ACM: HDU-1162 Eddy`s picture
*@Time: 18/9/14
*@IDE: VSCode + clang++
***********************************/
#include<cstdio >
#include<iostream>
#include<algorithm>
#include<queue>
#include<cmath>
using namespace std;
const int maxn = 110;
//坐标结构体
struct Node {
double x, y;//坐标
}node[maxn]; //结点集合
//边结构体
struct edge{
int u, v;
double cost;
edge() {}
edge(int _u, int _v, double _cost) : u(_u), v(_v), cost(_cost) {} //构造函数,便于加入结点
bool operator < (const edge& n) const { //规定优先级 - 浮点数比较
return cost - n.cost > 1e-8; //(注意和sort函数是相反的)
}
};
int N; //结点数
int far[maxn]; //并查集
//寻根
int find_root(int a) {
int root = a;
while(root != far[root]) root = far[root];
while(a != far[a]) { //路径压缩
int cur = a;
a = far[a];
far[cur] = root;
}
return root;
}
//合并集合
void union_set(int a, int b) {
int root_a = find_root(a);
int root_b = find_root(b);
if(a != b){
far[root_b] = root_a;
}
}
//Kruskal算法
double kruskal(priority_queue<edge> E) {
double ans = 0;//权值和
int edge_num = 0; //已选择的边数
for(int i = 0; i < E.size(); i++) {
edge e = E.top(); E.pop(); //get fisrt edge
int root_u = find_root(e.u);
int root_v = find_root(e.v);
if(root_u != root_v) { //if not belong to same set
union_set(root_u, root_v);
edge_num++;
ans += e.cost;
}
if(edge_num == N - 1) break; //边数==结点数-1
}
if(edge_num != N - 1) return -1;//边数==结点数-1
else return ans;
}//kruskal
//计算两点间的距离
double getDis(int a, int b) {
double x_dis = node[a].x - node[b].x;
double y_dis = node[a].y - node[b].y;
return sqrt(x_dis * x_dis + y_dis * y_dis); //勾股定理
}
int main() {
while(scanf("%d", &N) != EOF) {
for(int i = 1; i <= N; i++) far[i] = i; //初始化并查集
priority_queue<edge> E; //保存所有边(无clear()函数,每次重新定义时间最快)
//注:优先级设置和sort()函数是相反的
for(int i = 1; i <= N; i++) {
scanf("%lf %lf", &node[i].x, &node[i].y); //输入结点坐标
}
//暴力计算结点距离
for(int i = 1; i <= N; i++) {
for(int j = i + 1; j <= N; j++) {
double dis = getDis(i, j);
E.push(edge(i, j, dis));
//printf("connected: %d %d dis: %.1f\n", i , j, dis); //查看符合条件的边
}
}
double ans = kruskal(E);
if(ans == -1) printf("oh!\n");
else printf("%.2f\n", ans); //输出最小成本
}//while
system("pause");
return 0;
}