2024-02-01（最小生成树，二分图）-CSDN博客

如何证明当前这条边可以被选？

        假设不选当前边，得到了一棵树，然后将这条边加上，那么必然会出现一个环，在这个环上，一定可以找出一条不小于当前边的边，那么把当前边替换上去，结果一定不会变差。

1140. 最短网络 - AcWing题库 

Prim算法  裸题 

import java.util.*;public class Main{static int N = 110;static int n, res;static int[][] g = new int[N][N];static int[] dist = new int[N];//连通块外的其他点到连通块的最短距离static boolean[] st = new boolean[N];//是否在联通块内public static void prim(){Arrays.fill(dist, 0x3f3f3f3f);dist[1] = 0;res = 0;for(int i = 1; i <= n; i ++){int t = -1;for(int j = 1; j <= n; j ++){if(!st[j] && (t == -1 || dist[t] > dist[j])){t = j;}}if(dist[t] == 0x3f3f3f3f) return;//不连通，直接返回res += dist[t];st[t] = true;for(int j = 1; j <= n; j ++){//更新其他点到连通块的距离dist[j] = Math.min(dist[j], g[t][j]);}}System.out.print(res);return;}public static void main(String[] args){Scanner sc = new Scanner(System.in);n = sc.nextInt();for(int i = 1; i <= n; i ++){for(int j = 1; j <= n; j ++){g[i][j] = sc.nextInt();}}prim();}
}

 

1141. 局域网 - AcWing题库

相当于在这个图的每个连通块内的一棵最小生成树，相当于求原图的“最小生成森林”。 

  Kruskal算法：

1.将所有的边权从小到大排序

2.依次枚举每条边的a， b， w， 如果a，b不连通，那么就将当前边加入最小生成树中去。

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 c - o.c;}
}public class Main{static int N = 210;static int n, m, sum;static int[] p = new int[N];//并查集static PII[] q = new PII[N];public static int find(int x){if(p[x] != x) p[x] = find(p[x]);return p[x];//并查集的基本操作}public static int Kruskal(){Arrays.sort(q, 1, m + 1);int res = 0;for(int i = 1; i <= m; 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){//是否在同一个连通块中p[a] = b;res += c;//最小生成树的权重之和}}return res;}public static void main(String[] args){Scanner sc = new Scanner(System.in);n = sc.nextInt();m = sc.nextInt();for(int i = 1; i <= n; i ++) p[i] = i;for(int i = 1; i <= m; i ++){int a = sc.nextInt();int b = sc.nextInt();int c = sc.nextInt();q[i] = new PII(a, b, c);sum += c;}System.out.print(sum - Kruskal());}
}

 

1142. 繁忙的都市 - AcWing题库

普通最小生成树：权值之和最小

本题：最大权值最小 

 1.将所有边从小到大排序

 2.从小到大依次枚举每条边a，b，w

     如果a，b已经连通，直接pass

     如果a，b不连通，那么直接将这条边选出来

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 c - o.c;}
}public class Main{static int N = 310, M = 10010;static int n, m, res, cnt;static int[] p = new int[N];static PII[] q = new PII[M];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, 1, m + 1);for(int i = 1; i <= m; 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){p[a] = b;cnt ++;res = Math.max(res, c);}}System.out.print(cnt + " " + res);}public static void main(String[] args){Scanner sc = new Scanner(System.in);n = sc.nextInt();m = sc.nextInt();for(int i = 1; i <= n; i ++){p[i] = i;}for(int i = 1; i <= m; i ++){int a = sc.nextInt();int b = sc.nextInt();int c = sc.nextInt();q[i] = new PII(a, b, c);}Kruskal();}
}

 

1143. 联络员 - AcWing题库

//1.先将所有必选边加入并查集
//2.将所有非必选边从小到大排序
//3.依次枚举所有非必选边a，b，w
//  如果a，b联通，直接pass，如果不联通，就将当前边选上

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 c - o.c;}
}public class Main{static int N = 2010, M = 10010;static int n, m, res, k;static int[] p = new int[N];static PII[] q = new PII[M];public static int find(int x){if(p[x] != x) p[x] = find(p[x]);return p[x];}public static void Krustral(){Arrays.sort(q, 0, k);for(int i = 0; i < k; 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){p[a] = b;res += c;}}System.out.print(res);}public static void main(String[] args){Scanner sc = new Scanner(System.in);n = sc.nextInt();m = sc.nextInt();for(int i = 1; i <= n; i ++){p[i] = i;}for(int i = 1; i <= m; i ++){int t = sc.nextInt();int a = sc.nextInt();int b = sc.nextInt();int c = sc.nextInt();if(t == 1){p[find(a)] = find(b);res += c;}else{q[k ++] = new PII(a, b, c);}}Krustral();}
}

 

1144. 连接格点 - 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 c - o.c;}
}public class Main{static int N = 1010, M = N * N, K = 2 * M;static int n, m, k, res;static PII[] q = new PII[K];static int[] p = new int[M];static int[][] g = new int[N][N];public static int find(int x){if(x != p[x]) p[x] = find(p[x]);return p[x];}public static void get_PII(){int[] dx = {-1, 0, 1, 0}, dy = {0, 1, 0, -1}, dw = {1, 2, 1, 2};for(int z = 0; z < 2; z ++){for(int i = 1; i <= n; i ++){for(int j = 1; j <= m; j ++){for(int u = 0; u < 4; u ++){if(u % 2 == z){int x = i + dx[u], y = j + dy[u];if(x <= 0 || y <= 0 || x > n || y > m) continue;int a = g[i][j], b = g[x][y], w = dw[u];if(a < b) q[k ++] = new PII(a, b, w);}}}}}}public static void main(String[] args){Scanner sc = new Scanner(System.in);n = sc.nextInt();m = sc.nextInt();for(int i = 1; i <= n * m; i ++){p[i] = i;}int cnt = 0;for(int i = 1; i <= n; i ++){//给每个点编号for(int j = 1; j <= m; j ++){g[i][j] = ++ cnt;}}get_PII();while(sc.hasNext()){int x1 = sc.nextInt();int y1 = sc.nextInt();int x2 = sc.nextInt();int y2 = sc.nextInt();int a = g[x1][y1], b = g[x2][y2];p[find(a)] = find(b);}for(int i = 0; i < k; 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){p[a] = b;res += c;}}System.out.print(res);}
}