PAT（甲）1003. Emergency（25）

原题连接：

1003. Emergency

我的思路:

利用 DFS（深度优先搜索）从起点开始找路（同时记录路径长度和经过的各个节点）

每找到一条可达路径，先判定这条路径的长度

如果大于当前最短路则直接跳过

如果等于当前最短路，则查看该路上可以集结几只队伍，如果队伍数大于当前最多队伍数，则更新，否则跳过

如果小于当前最短路，则更新当前最短路，并更新当前最多队伍

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {
private static List<Integer> path = new ArrayList<>();
private static int minLength = 0, maxTeams = 0, thisLength = 0, viablePath = 0;
private static int[] teams;
private static boolean[] marked;

public static void main(String[] args) {
Scanner in = new Scanner(System.in);

int[][] cityMap;  //我使用邻接矩阵法表示图

int n = in.nextInt();
int m = in.nextInt();
int from = in.nextInt();
int to = in.nextInt();

teams = new int
;
marked = new boolean
;
cityMap = new int

;

for (int i = 0; i < n; i++){
teams[i] = in.nextInt();
}

/*
注意：在构造图时, 因为我使用邻接矩阵法表示图, 而且图本身为无向图
所以在构造时 start->end 和 end->start 都要赋值
*/
for (int i = 0; i < m; i++){
int start, end, len;
start = in.nextInt();
end = in.nextInt();
len = in.nextInt();
cityMap[start][end] = len;
cityMap[end][start] = len;
}

dfs(cityMap, from, to);
System.out.println(viablePath + " " + (maxTeams + teams[from]));
}

private static void dfs(int[][] map, int from, int to){
if (from == to) {
if (minLength <= 0 || thisLength <= minLength){
if (thisLength == minLength){
viablePath++;
}else {
minLength = thisLength;
viablePath = 1;
maxTeams = 0;
}
int thisTeams = 0;
for (int i = 0; i < path.size(); i++){
thisTeams += teams[path.get(i)];
}
if (thisTeams > maxTeams){
maxTeams = thisTeams;
}
}
return;
}

/*
使用 dfs 找路
注意：因为我们不是遍历图, 而是寻找所有的路径
所以要在回溯时及时恢复各个变量的原值
*/
marked[from] = true;
for (int i = 0; i < map.length; i++){
if (map[from][i] > 0 && !marked[i]){
path.add(i);
thisLength += map[from][i];
dfs(map, i, to);
thisLength -= map[from][i];
path.remove(path.size() - 1);
}
}
marked[from] = false;
}
}