HDOJ 1162:Eddy's picture 求解最小生成树

来源：互联网发布：war包和源码区别编辑：程序博客网时间：2024/05/22 01:40

http://acm.hdu.edu.cn/showproblem.php?pid=1162

这道题也是一道单纯的求解最小生成树的题目，是个稠密图，所以是普里姆算法的用武之地。对于完全图，普里姆算法的复杂度（n^2）,而克鲁斯卡尔算法的复杂度（n^2logn^2），效率大小长差很大。

我的AC代码：

#include <iostream>#include <stdio.h>#include <numeric>#include <math.h>using namespace std;const int Max = 110;double g[Max][Max], x[Max], y[Max];int vers;struct MST{double dist;bool belong;}Mst[Max];void Prim(){Mst[1].belong = true;for(int i=2; i<=vers; ++i){Mst[i].dist = g[1][i];Mst[i].belong = false;}for(int i=2; i<=vers; i++){int pos = 1;while(Mst[pos].belong) ++pos;for(int j=pos+1; j<=vers; ++j)if(!Mst[j].belong && Mst[j].dist < Mst[pos].dist) pos = j;Mst[pos].belong = true;for(int j=1; j<=vers; ++j)if(!Mst[j].belong && g[pos][j] < Mst[j].dist) Mst[j].dist = g[pos][j];}}double distance(double x1, double y1, double x2, double y2){return sqrt((y2-y1) * (y2-y1) + (x2-x1) * (x2-x1));}double add(double a, MST &b){return a + b.dist; }int main(){double sum;while(scanf("%d", &vers) != EOF){for(int i=1; i<=vers; ++i) scanf("%lf %lf", x+i, y+i);for(int i=1; i<=vers; ++i)for(int j=1; j<=vers; ++j)g[i][j]= distance(x[i], y[i], x[j], y[j]);Prim();sum = accumulate(Mst+1, Mst+vers+1, 0.0, add);printf("%.2lf\n", sum);}system("pause");return 0;}