1146. 新的开始 - AcWing题库
//建立一个虚拟远点
import java.util.*;public class Main{static int N = 310;static int[][] w = new int[N][N];static int[] dist = new int[N];static boolean[] st = new boolean[N];static int n, res;public static int prim(){Arrays.fill(dist, 0x3f3f3f3f);dist[0] = 0;for(int i = 0; i < n + 1; i ++){int t = -1;for(int j = 0; j < n + 1; j ++){if(!st[j] && (t == -1 || dist[t] > dist[j])){t = j;}}st[t] = true;res += dist[t];for(int j = 0; j < n + 1; j ++){dist[j] = Math.min(dist[j], w[t][j]);//t点到这个点的距离短还是虚拟远点到这个点的距离短}}return res;}public static void main(String[] args){Scanner sc = new Scanner(System.in);n = sc.nextInt();for(int i = 1; i <= n; i ++){//虚拟远点到每个点的距离w[0][i] = sc.nextInt();w[i][0] = w[0][i];}for(int i = 1; i <= n; i ++){for(int j = 1; j <= n; j ++){w[i][j] = sc.nextInt();}}System.out.print(prim());}
}
1145. 北极通讯网络 - AcWing题库
找到一个最小的d值,将所有权值大于d的边删去,整个图形连通块的数量不超过k条
import java.util.*;class PII_1{int x, y;public PII_1(int x, int y){this.x = x;this.y = y;}
}class PII_2 implements Comparable<PII_2>{int a, b;double c;public PII_2(int a, int b, double c){this.a = a;this.b = b;this.c = c;}public int compareTo(PII_2 o){return Double.compare(c, o.c);}
}public class Main{static int N = 510, M = N * N;static PII_1[] a = new PII_1[M];static PII_2[] q = new PII_2[M];static int[] p = new int[M];static int n, k, m;public static double get_dist(PII_1 a, PII_1 b){int dx = a.x - b.x;int dy = a.y - b.y;return Math.sqrt(dx * dx + dy * dy);}public static int find(int x){if(p[x] != x) p[x] = find(p[x]);return p[x];}public static void Kruskal(){Arrays.sort(q, 0, m);//一定要记得排序int cnt = n;double res = 0.0;for(int i = 0; i < m; i ++){if(cnt <= k) break;int a = q[i].a;int b = q[i].b;double c = q[i].c;a = find(a);b = find(b);if(a != b){p[a] = b;cnt --;res = c;}}System.out.printf("%.2f", res);}public static void main(String[] args){Scanner sc = new Scanner(System.in);n = sc.nextInt();k = sc.nextInt();for(int i = 0; i < n; i ++) p[i] = i;for(int i = 0; i < n; i ++){int x = sc.nextInt();int y = sc.nextInt();a[i] = new PII_1(x, y);//每个村庄的坐标}for(int i = 0; i < n; i ++){for(int j = 0; j < n; j ++){//村庄与村庄之间两两的距离q[m ++] = new PII_2(i, j, get_dist(a[i], a[j]));}}Kruskal();}
}
346. 走廊泼水节 - AcWing题库
import java.util.*;class PII implements Comparable<PII>{int a, b, c;public PII(int a, int b, int c){this.a = a;this.b = b;this.c = c;}public int compareTo(PII o){return Integer.compare(c, o.c);}
}public class Main{static int N = 6010, n;static int[] p = new int[N];static int[] size = new int[N];static PII[] q = new PII[N];public static int find(int x){if(p[x] != x) p[x] = find(p[x]);return p[x];}public static void Kruskal(){Arrays.sort(q, 0, n - 1);int res = 0;for(int i = 0; i < n - 1; i ++){int a = q[i].a;int b = q[i].b;int c = q[i].c;a = find(a);b = find(b);if(a != b){res += (size[a] * size[b] - 1) * (c + 1);//权值加1p[a] = b;size[b] += size[a];}}System.out.println(res);}public static void main(String[] args){Scanner sc = new Scanner(System.in);int T = sc.nextInt();while(T -- > 0){n = sc.nextInt();for(int i = 1; i <= n; i ++){p[i] = i;size[i] = 1;}for(int i = 0; i < n - 1; i ++){int a = sc.nextInt();int b = sc.nextInt();int c = sc.nextInt();q[i] = new PII(a, b, c);//先把点加进去}Kruskal();}}
}
1148. 秘密的牛奶运输 - AcWing题库
次小生成树:
定义:给一个带权的图,把图的所有生成树按权值从小到大排序,第二小的称为次小生成树(有些题目最小生成树不唯一,次小生成树可以和最小生成树相等)
方法一:先求最小生成树,然后依次枚举删去最小生成树的边求解O(mlogm + nm)
只能求出来非严格最小生成树。
方法二:先求最小生成树,然后依次枚举非树边,将该边加入到树中,同时从树中去掉一条 边,使得最终的图仍是一棵树,则一定可以求出次小生成树。