JAVA CCF-201609-4 交通规划

欢迎访问我的CCF认证解题目录

题目描述

思路过程

最短路径的变形题，首先要确保点到首都的最短距离不变，再去节省修路的费用，我们可以定义一个数组pre保存上一个节点，当访问length[index]时，length[pre[index]]一定是已经访问过的了，修建时用sum += length[index] - length[pre[index]];去计算。

代码

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;

public class Main {
	
	public static void main(String[] args) throws IOException {
		BufferedReader by = new BufferedReader(new InputStreamReader(System.in));
		String[] line = by.readLine().split(" ");//读入行
		int n = Integer.parseInt(line[0]), m = Integer.parseInt(line[1]);
		long INF = Long.MAX_VALUE/2-1;//无限大
		long sum = 0;
		int[] pre = new int[n+1];//存储节点的上一节点
		ArrayList<ArrayList<Node>> graph = new ArrayList<ArrayList<Node>>();
		boolean[] flag = new boolean[n+1];
		long[] length = new long[n+1];
		for ( int i = 0; i <= n; i++ ) {
			pre[i] = 1;//初始为1，首都
			length[i] = INF;
			graph.add(new ArrayList<Node>());
		}
		while ( m-- != 0 ) {
			line = by.readLine().split(" ");
			graph.get(Integer.parseInt(line[0])).add(new Node(Integer.parseInt(line[1]), Integer.parseInt(line[2])));
			graph.get(Integer.parseInt(line[1])).add(new Node(Integer.parseInt(line[0]), Integer.parseInt(line[2])));
		}
		
		
		length[1] = 0;
		while ( true ) {
			int index = -1;
			long min = INF;
			
			//找最近的点
			for ( int i = 1; i <= n; i++ ) {
				if ( !flag[i] && length[i] < min ) {
					min = length[i];
					index = i;
				}
			}
			
			if ( index == -1 ) break;
			flag[index] = true;
			sum += length[index] - length[pre[index]];//要修建的钱
			
			//优化
			for ( Node node: graph.get(index) ) {
				if ( length[node.v] >= length[index] + node.length ) {
					length[node.v] = length[index] + node.length;
					if ( length[index] > length[pre[node.v]] ) {//如果能让后面的修建更短
						pre[node.v] = index;
					}
				}
			}
		}
		
		System.out.println(sum);
	}
	
}
class Node {
	int v;//点
	long length;//长度
	public Node(int v, long length) {
		this.v = v;
		this.length = length;
	}
	
}