Problem: AtCoder ABC 237 E — Skiing · 한국어 · 日本語
For a route from vertex 1 to a vertex v, let U be the total amount climbed and D the total amount descended. The happiness change along the route is D - 2U: descending by a unit gains one, while climbing by a unit loses two. Since the net height change is H[v] - H[1] = U - D, we have D = U + H[1] - H[v], so the happiness is H[1] - H[v] - U. For a fixed destination, its height is fixed, so maximizing happiness is equivalent to minimizing the total climb U.
Assign each undirected road u-v a directed cost in each direction: going from u to v costs max(0, H[v] - H[u]), the climb on that step. Going from v to u similarly costs max(0, H[u] - H[v]). Descending and equal-height steps cost zero. All costs are nonnegative, so Dijkstra’s algorithm finds the minimum accumulated climb from vertex 1 to every reachable vertex. The edge costs are directed even though the roads are undirected; in particular, an equal-height road has zero cost in both directions.
For each reachable vertex v, compute H[1] - H[v] - dist[v], where dist[v] is the minimum climb. Unreachable vertices have no route from the start and must not be included. Initialize the answer to zero because vertex 1 is reachable from itself with happiness zero. Distances, height differences, and happiness are stored as long to avoid overflow when adding costs. The running time is O((N + M) log N) and the space usage is O(N + M).
Java
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import java.io.BufferedInputStream;
import java.io.IOException;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.PriorityQueue;
public class Main {
private static class Edge {
int to;
long climb;
Edge(int to, long climb) {
this.to = to;
this.climb = climb;
}
}
private static class State implements Comparable<State> {
int vertex;
long distance;
State(int vertex, long distance) {
this.vertex = vertex;
this.distance = distance;
}
@Override
public int compareTo(State other) {
return Long.compare(distance, other.distance);
}
}
private static class FastScanner {
private final BufferedInputStream input = new BufferedInputStream(System.in);
private final byte[] buffer = new byte[1 << 16];
private int length;
private int position;
private int read() throws IOException {
if (position == length) {
length = input.read(buffer);
position = 0;
if (length == -1) {
return -1;
}
}
return buffer[position++];
}
long nextLong() throws IOException {
int c;
do {
c = read();
} while (c <= ' ' && c != -1);
long value = 0;
while (c > ' ') {
value = value * 10 + c - '0';
c = read();
}
return value;
}
}
public static void main(String[] args) throws IOException {
FastScanner scanner = new FastScanner();
int n = (int) scanner.nextLong();
int m = (int) scanner.nextLong();
long[] height = new long[n];
for (int i = 0; i < n; i++) {
height[i] = scanner.nextLong();
}
List<List<Edge>> graph = new ArrayList<>(n);
for (int i = 0; i < n; i++) {
graph.add(new ArrayList<>());
}
for (int i = 0; i < m; i++) {
int u = (int) scanner.nextLong() - 1;
int v = (int) scanner.nextLong() - 1;
graph.get(u).add(new Edge(v, Math.max(0L, height[v] - height[u])));
graph.get(v).add(new Edge(u, Math.max(0L, height[u] - height[v])));
}
long[] distance = new long[n];
Arrays.fill(distance, Long.MAX_VALUE);
distance[0] = 0;
PriorityQueue<State> queue = new PriorityQueue<>();
queue.add(new State(0, 0));
while (!queue.isEmpty()) {
State current = queue.poll();
if (current.distance != distance[current.vertex]) {
continue;
}
for (Edge edge : graph.get(current.vertex)) {
long nextDistance = current.distance + edge.climb;
if (nextDistance < distance[edge.to]) {
distance[edge.to] = nextDistance;
queue.add(new State(edge.to, nextDistance));
}
}
}
long answer = 0;
for (int v = 0; v < n; v++) {
if (distance[v] != Long.MAX_VALUE) {
answer = Math.max(answer, height[0] - height[v] - distance[v]);
}
}
System.out.println(answer);
}
}