第一题,求最大得分和最大得分人数,以及他们的下标。按照题意求解,比较找最大值,输出即可。
第二题,无向图中求不舒服的路径。保证无环,边的权重也就是不舒服的程度,求a到b的路径中的不舒服的最大值。
比如:
5 4 3 // 节点数、边数、待求解路径个数 1 2 1 // 节点1 与 节点2 之间的不舒服程度为1 2 3 2 3 4 1 4 5 6 2 5 // 待求解 节点2 到 节点 5 4 1 5 3
输出:626
可以采用dfs来遍历。但昨天没做出来,通过18%。刚刚写了下,发现是回溯的时候边界条件的问题。记录下:
import java.util.*; import java.util.Scanner; public class TestC { private static void insertData(Map<Integer, List<NodeT>> graph, int a, int b, int c){ if(!graph.containsKey(a)){ ArrayList<NodeT> lists = new ArrayList<>(); lists.add(new NodeT(b, c)); graph.put(a, lists); }else{ List<NodeT> list = graph.get(a); list.add(new NodeT(b, c)); graph.put(a, list); } } public static void main(String[] args) { Scanner scanner = new Scanner(System.in); String[] lines = scanner.nextLine().split(" "); int vectorNumber = Integer.parseInt(lines[0]); int edgeNumber = Integer.parseInt(lines[1]); int questionNumber = Integer.parseInt(lines[2]); Map<Integer, List<NodeT>> graph = new HashMap<>(); Map<String, Integer> weight = new HashMap<>(); for (int i = 0; i < edgeNumber; i++) { String[] temp = scanner.nextLine().split(" "); int a = Integer.parseInt(temp[0]); int b = Integer.parseInt(temp[1]); int c = Integer.parseInt(temp[2]); insertData(graph, a, b, c); insertData(graph, b, a, c); weight.put(a+"to"+b, c); weight.put(b+"to"+a, c); } List<Integer> resu = new ArrayList<>(); for (int i = 0; i < questionNumber; i++) { String[] temp = scanner.nextLine().split(" "); int a = Integer.parseInt(temp[0]); int b = Integer.parseInt(temp[1]); if(a == b){ resu.add(0); }else{ List<Integer> res = new ArrayList<>(); List<List<Integer>> tempt = new ArrayList<>(); res.add(a); boolean[] visited = new boolean[vectorNumber + 1]; visited[a] = true; dfs(graph, a, b, res, visited, tempt); res = tempt.get(0); int w = 0; for (int j = 1; j < res.size(); j++) { w = Math.max(w, weight.get(res.get(j - 1) + "to" + res.get(j))); } resu.add(w); } } for (Integer integer : resu) { System.out.println(integer); } } private static void dfs(Map<Integer, List<NodeT>> graph, int a, int b, List<Integer> temp, boolean[] visited, List<List<Integer>> res) { List<NodeT> neighbors = graph.get(a); for (NodeT node : neighbors) { if(!visited[node.n]){ visited[node.n] = true; temp.add(node.n); if (node.n == b) { res.add(new ArrayList<>(temp)); return; } dfs(graph, node.n, b, temp, visited, res); temp.remove(temp.size() - 1); visited[node.n] = false; } } } static class NodeT{ int n; int weight; public NodeT(int n, int weight){ this.n = n; this.weight = weight; } @Override public boolean equals(Object o) { if (this == o) return true; if (o == null || getClass() != o.getClass()) return false; NodeT nodeT = (NodeT) o; return n == nodeT.n && weight == nodeT.weight; } @Override public int hashCode() { return Objects.hash(n, weight); } } /** 5 4 3 1 2 1 2 3 2 3 4 1 4 5 6 2 5 4 1 5 3 */ }
全部评论
(1) 回帖