计算给定社交网络中某个用户在k跳范围内的影响力,也就是要找到所有距离m,k以内的能访问到的点,最短路问题,跑一遍dijstra,之后,枚举所有点,距离小于等于k的ans+=1,最后输出ans-1(除去本身)即可
#include <bits/stdc++.h>
using namespace std;
#define N 200005
int vis[N];
int dis[N];
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