题意：

用Dijkstra来做，题目在Dijkstra的基础上增加了一些操作，除了要更新dist[] 和 visit[] 数组外，将原算法中记录路径的parent[]数组改为--->记录start至end最短路径上的rescue[]，另外还需要一个ways[]数组来更新最短路径的条数。在遇到多条最短路径时，表现为dist[j] == dist[index] + edge[index][j]，此时将ways[j] += ways[index]即可。然后比较rescue[j] 与 rescue[index] + weight[j]的大小确定rescue[j]始终是到达j点的所有最短路径中最大的。

import java.util.Scanner;
public class Main {
	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);
		int n = input.nextInt();
		int m = input.nextInt();
		int c1 = input.nextInt();
		int c2 = input.nextInt();
		
		int i=0, j=0;
		
		int weight[] = new int[n];
		for(i=0; i<weight.length; i++)
			weight[i] = input.nextInt();
		
		int edge[][] = new int[n][n];
		for(i=0; i<n; i++)
			for(j=0; j<n; j++) 
				if(i != j) edge[i][j] = Integer.MAX_VALUE;
		for(i=0; i<m; i++) { 
			int a = input.nextInt();
			int b = input.nextInt();
			edge[a][b] = edge[b][a] = input.nextInt();
		}
			
		
		int dist[] = new int[n];
		for(i=0; i<n; i++)
			dist[i] = Integer.MAX_VALUE;
		
		boolean visit[] = new boolean[n];
		int rescue[] = new int[n];
		int ways[] = new int[n];
		
		//dijistra
		dist[c1] = 0;
		rescue[c1] = weight[c1];
		ways[c1] = 1;
		
		int k=0;
		
		for(i=0; i<n; i++) {
			int index = -1, min = Integer.MAX_VALUE;
			for(k=0; k<n; k++) {
				if(!visit[k] && min > dist[k]) {
					index = k;
					min = dist[k];
				}
			}
			if(index == -1) break;
			visit[index] = true;
			
			for(j=0; j<n; j++) {
				if(!visit[j] && edge[index][j] != Integer.MAX_VALUE) {
					if(dist[index] + edge[index][j] < dist[j]) {
						dist[j] = dist[index] + edge[index][j];
						rescue[j] = rescue[index] + weight[j];
						ways[j] = ways[index];
					}
					else if(dist[index] + edge[index][j] == dist[j]){	
						ways[j] += ways[index];
						if(rescue[j] < rescue[index] + weight[j])
							rescue[j] = rescue[index] + weight[j];
					}
				}
			}
		}
		System.out.print(ways[c2] + " " + rescue[c2]);
	}
}

标签：25,dist,rescue,Emergency,index,int,nextInt,input,1003
来源： https://blog.csdn.net/LangWeiXian_/article/details/114396285