题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=1162
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10085 Accepted Submission(s):
5094
19975116 2017-03-03 08:53:42 Accepted 1162 0MS 1524K 1304 B G++ #include <stdio.h> #include <math.h> #include <string.h> #include <algorithm> using namespace std; struct point { int x,y; double l; }p[11000]; int parent[110],n; double x[110],y[110]; bool cmp(point a, point b) { return a.l < b.l; } int find(int x) { int s,tmp; for (s = x; parent[s] >= 0; s = parent[s]); while (s != x) { tmp = parent[x]; parent[x] = s; x = tmp; } return s; } void Union(int A, int B) { int a = find(A), b = find(B); int tmp = parent[a]+parent[b]; if (parent[a] < parent[b]) { parent[b] = a; parent[a] = tmp; } else { parent[a] = b; parent[b] = tmp; } } void kruskal(int k) { double sum = 0; int u,v,i,j = 0; memset(parent,-1,sizeof(parent)); for (i = 0; i <= k; i ++) { u = p[i].x; v = p[i].y; if (find(u) != find(v)) { sum += p[i].l; Union(u,v); j ++; } if (j == n-1) break; } printf("%.2lf\n",sum); } int main () { int i,j,k; while (scanf("%d",&n)!=EOF) { k = -1; for (i = 1; i <= n; i ++) scanf("%lf%lf",x+i,y+i); for (i = 1; i < n; i ++) for (j = 1+i; j <= n; j ++) { p[++ k].x = i; p[k].y = j; p[k].l = sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j])); } sort(p,p+k+1,cmp); kruskal(k); } return 0; }
hdu 1162 Eddy's picture (Kruskal 算法)
原文:http://www.cnblogs.com/yoke/p/6506455.html