1334. Find the City With the Smallest Number of Neighbors at a Threshold Distance

Source: https://leetcode.com/contest/weekly-contest-173/problems/find-the-city-with-the-smallest-number-of-neighbors-at-a-threshold-distance/

There are n cities numbered from 0 to n-1. Given the array edges where edges[i] = [fromi, toi, weighti] represents a bidirectional and weighted edge between cities fromi and toi, and given the integer distanceThreshold.

Return the city with the smallest number of cities that are reachable through some path and whose distance is at most distanceThreshold, If there are multiple such cities, return the city with the greatest number.

Notice that the distance of a path connecting cities i and j is equal to the sum of the edges’ weights along that path.

Example 1:

Input: n = 4, edges = [[0,1,3],[1,2,1],[1,3,4],[2,3,1]], distanceThreshold = 4
Output: 3
Explanation: The figure above describes the graph.
The neighboring cities at a distanceThreshold = 4 for each city are:
City 0 -> [City 1, City 2]
City 1 -> [City 0, City 2, City 3]
City 2 -> [City 0, City 1, City 3]
City 3 -> [City 1, City 2]
Cities 0 and 3 have 2 neighboring cities at a distanceThreshold = 4, but we have to return city 3 since it has the greatest number.

Example 2:

Input: n = 5, edges = [[0,1,2],[0,4,8],[1,2,3],[1,4,2],[2,3,1],[3,4,1]], distanceThreshold = 2
Output: 0
Explanation: The figure above describes the graph.
The neighboring cities at a distanceThreshold = 2 for each city are:
City 0 -> [City 1]
City 1 -> [City 0, City 4]
City 2 -> [City 3, City 4]
City 3 -> [City 2, City 4]
City 4 -> [City 1, City 2, City 3]
The city 0 has 1 neighboring city at a distanceThreshold = 2.

Constraints:

2 <= n <= 100
1 <= edges.length <= n * (n - 1) / 2
edges[i].length == 3
0 <= fromi < toi < n
1 <= weighti, distanceThreshold <= 10^4
All pairs (fromi, toi) are distinct.

思路：求出所有点之间最短路径长度，从而计算从各顶点出发可到达的distanceThreshold以内的点的数量，找出最大的即可，一样大的也更新，因为要下标大的。

class Solution {
public:
    int adjmat[100][100];
    int dist[100][100];
    int mark[100];
    int num[100];
    void computedist(int n, int s){
        memset(mark,0,sizeof(mark));
        mark[s]=1;
        for(int i=0;i<n;i++){
            dist[s][i]=adjmat[s][i];
        }
        dist[s][s]=0;
        
        for(int k=0;k<n;k++){
            int mindist=0x3f3f3f3f;
            int minidx;
            for(int i=0;i<n;i++){
                if(dist[s][i]!=0x3f3f3f3f && mark[i]==0 && dist[s][i]<mindist){
                    mindist=dist[s][i];
                    minidx=i;
                }
            }
            mark[minidx]=1;
            for(int i=0;i<n;i++){
                if(adjmat[minidx][i]!=0x3f3f3f3f && mark[i]==0 && adjmat[minidx][i]+dist[s][minidx]<dist[s][i]){
                    dist[s][i]=adjmat[minidx][i]+dist[s][minidx];
                }
            }
            
        }
    }
    int findTheCity(int n, vector<vector<int>>& edges, int distanceThreshold) {
        int numedge=edges.size();
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                adjmat[i][j]=0x3f3f3f3f;
            }
        }
        for(int i=0;i<numedge;i++){
            adjmat[edges[i][0]][edges[i][1]]=edges[i][2];
            adjmat[edges[i][1]][edges[i][0]]=edges[i][2];
        }
        memset(num,0,sizeof(num));
        int minnum=105;
        int ans;
        for(int i=0;i<n;i++){
            computedist(n,i);
            for(int j=0;j<n;j++){
                if(dist[i][j]<=distanceThreshold){
                    num[i]++;
                }
            }
            if(num[i]<=minnum){
                minnum=num[i];
                ans=i;
            }
        }
        
        return ans;
    }
};