【解決方法】無向グラフに関するこの質問に役立つ可能性のある疑似 CDE またはアルゴリズムはありますか?

You are given a city network in the form of a tree(an acyclic graph). There are N cities and each city has a unique ID (integer from 1 to N). There exists a unique path between each pair of cities. There are only N-1 paths in the city network. The population of each city is provided.

Given Q queries, you need return the number of cities that have a population of at most W on the path from U to V. U and V are the given city ID.

Inputs:
No. of cities = 5

Population for each cities = 4 7 8 6 4 

Road Available: 1 and 5, 
                5 and 2, 
                2 and 3,
                2 and 4

No. of Queries: 5

Query 1(U,V,W): 4 2 1

Query 2(U,V,W): 2 3 7

Query 3(U,V,W): 2 4 8

Query 4(U,V,W): 5 3 1

Query 5(U,V,W): 3 4 8


Expected output: 0
                 1  
                 2
                 0
                 3
```
#include<stdio.h>
#include <stdlib.h>
  
int* city_population (int N, int* population, int** road, int Q, int** cities) ;

int main() {
    int N;
    scanf("%d", &N);
    int i_population;
    int *population = (int *)malloc(sizeof(int)*(N));
    for(i_population = 0; i_population < N; i_population++)
    	scanf("%d", &population[i_population]);
    int i_road, j_road;
    int **road = (int **)malloc((N-1)*sizeof(int *));
    for(i_road = 0; i_road < N-1; i_road++)
    {
    	road[i_road] = (int *)malloc((2)*sizeof(int));
    }
    for(i_road = 0; i_road < N-1; i_road++)
    {
    	for(j_road = 0; j_road < 2; j_road++)
    	{
    		scanf("%d", &road[i_road][j_road]);
    	}
    }
    int Q;
    scanf("%d", &Q);
    int i_cities, j_cities;
    int **cities = (int **)malloc((Q)*sizeof(int *));
    for(i_cities = 0; i_cities < Q; i_cities++)
    {
    	cities[i_cities] = (int *)malloc((3)*sizeof(int));
    }
    for(i_cities = 0; i_cities < Q; i_cities++)
    {
    	for(j_cities = 0; j_cities < 3; j_cities++)
    	{
    		scanf("%d", &cities[i_cities][j_cities]);
    	}
    }

    int* out_ = city_population(N, population, road, Q, cities);
    printf("%d", out_[0]);
    int i_out_;
    for(i_out_ = 1; i_out_ < Q; i_out_++)
    	printf("\n%d", out_[i_out_]);
}
int* city_population (int N, int* population, int** road, int Q, int** cities) 
{


}

“`
この質問に役立つアルゴリズムの疑似コードはありますか?

私が試したこと：

利用できるスタックやキューがないので、どこから始めればよいかわかりません