2024-01-31（最短路径）-CSDN博客

   求负环的常用方法，基于spfa：

1.统计每个点入队的次数，如果有个点入队n次，则说明存在负环

2.统计当前每个点的最短路中包含的边数，如果某个点的最短路的所包含的边数大于等于         n，则说明也存在环 

当所有点的入队次数大于2n时，我们就认为图中有很大的可能是存在负环的 

(如果TLE了，那么可以把队列换成栈来试一下）

904. 虫洞 - AcWing题库

裸题 

import java.util.*;public class Main{static int N = 510, M = 5500;static int n, m1, m2, idx;static int[] h = new int[N], e = new int[M], ne = new int[M], w = new int[M];static int[] dist = new int[N];static boolean[] st = new boolean[N];static int[] cnt = new int[M];//边数public static void add(int a, int b, int c){e[idx] = b;w[idx] = c;ne[idx] = h[a];h[a] = idx ++;}public static boolean spfa(){Arrays.fill(dist, 0);Arrays.fill(st, false);Arrays.fill(cnt, 0);Queue<Integer> q = new LinkedList<>();//循环队列for(int i = 1; i <= n; i ++){//每个点入队q.offer(i);st[i] = true;}while(!q.isEmpty()){int t = q.poll();st[t] = false;for(int i = h[t]; i != -1; i = ne[i]){int j = e[i];if(dist[j] > dist[t] + w[i]){dist[j] = dist[t] + w[i];cnt[j] = cnt[t] + 1;if(cnt[j] >= n) return true;if(!st[j]){q.offer(j);st[j] = true;}}}}return false;}public static void main(String[] args){Scanner sc = new Scanner(System.in);int T = sc.nextInt();while(T -- > 0){n = sc.nextInt();m1 = sc.nextInt();m2 = sc.nextInt();Arrays.fill(h, -1);idx = 0;//一定要记得！！！while(m1 -- > 0){int a = sc.nextInt();int b = sc.nextInt();int c = sc.nextInt();add(a, b, c);add(b, a, c);}while(m2 -- > 0){int a = sc.nextInt();int b = sc.nextInt();int c = sc.nextInt();add(a, b, -c);//虫洞的通道是单向的}if(spfa()) System.out.println("YES");else System.out.println("NO");}}
}

361. 观光奶牛 - AcWing题库

01分数规划：二分 -> 整理不等式 -> 重新定义边权 -> 判断图中是否存在正环

import java.util.*;public class Main{static int N = 1010, M = 5010;static int n, m, idx;static int[] wd = new int[N];//点权static int[] h = new int[N], e = new int[M], ne = new int[M], wb = new int[M];//边权static double[] dist = new double[N];static int[] cnt = new int[N];static boolean[] st = new boolean[N];public static void add(int a, int b, int c){e[idx] = b;wb[idx] = c;ne[idx] = h[a];h[a] = idx ++;}public static boolean spfa(double x){Arrays.fill(dist, 0);Arrays.fill(st, false);Arrays.fill(cnt, 0);Queue<Integer> q = new LinkedList<>();for(int i = 1; i <= n; i ++){q.offer(i);st[i] = true;}while(!q.isEmpty()){int t = q.poll();st[t] = false;for(int i = h[t]; i != -1; i = ne[i]){int j = e[i];if(dist[j] < dist[t] + wd[t] - x * wb[i]){dist[j] = dist[t] + wd[t] - x * wb[i];cnt[j] = cnt[t] + 1;if(cnt[j] >= n) return true;if(!st[j]){q.offer(j);st[j] = true;}}}}return false;}public static void main(String[] args){Scanner sc = new Scanner(System.in);n = sc.nextInt();m = sc.nextInt();Arrays.fill(h, -1);for(int i = 1; i <= n; i ++){wd[i] = sc.nextInt();}for(int i = 0; i < m;  i ++){int a = sc.nextInt();int b = sc.nextInt();int c = sc.nextInt();add(a, b, c);}//二分double l = 0, r = 1010;while(r - l > 1e-4){double mid = (l + r) / 2;if(spfa(mid)) l = mid;else r = mid;}System.out.printf("%.2f", l);}
}

1165. 单词环 - AcWing题库

这道题就用到了前面说的技巧 

import java.util.*;public class Main{static int N = 700, M = 100010;static int n, idx;static int[] h = new int[N], e = new int[M], ne = new int[M], w = new int[M];static int[] cnt = new int[N];static boolean[] st = new boolean[N];static double[] dist = new double[N];public static void add(int a, int b, int c){e[idx] = b;w[idx] = c;ne[idx] = h[a];h[a] = idx ++;}public static boolean check(double x){Arrays.fill(st, false);Arrays.fill(dist, 0);Arrays.fill(cnt, 0);int count = 0;Queue<Integer> q = new LinkedList<>();for(int i = 0; i < 676; i ++){q.offer(i);st[i] = true;}while(!q.isEmpty()){int t = q.poll();st[t] = false;for(int i = h[t]; i != -1; i = ne[i]){int j = e[i];if(dist[j] < dist[t] + w[i] - x){dist[j] = dist[t] + w[i] - x;cnt[j] = cnt[t] + 1;if(++ count > 10000) return true;if(cnt[j] >= N) return true;if(!st[j]){q.offer(j);st[j] = true;}}}}return false;}public static void main(String[] args){Scanner sc = new Scanner(System.in);while(true){n = sc.nextInt();if(n == 0) break;Arrays.fill(h, -1);idx = 0;for(int i = 0; i < n; i ++){String str = sc.next();int len = str.length();if(len < 2) continue;//转化为26进制数int left = (str.charAt(0) - 'a') * 26 + (str.charAt(1) - 'a');int right = (str.charAt(len - 2) - 'a') * 26 + (str.charAt(len - 1) - 'a');add(left, right, len);}if(!check(0)){System.out.println("No solution");continue;}else{double l = 0, r = 1000;while(r - l > 1e-4){double mid = (l + r) / 2;if(check(mid)) l = mid;else r = mid;}System.out.printf("%.2f\n", r);}}}
}