#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int N = 100010;
int n, m;
int h[N], e[N], ne[N], idx;
int d[N],q[N];
int ans = N;
void add(int a, int b)
{
e[idx] = b, ne[idx] = h[a], h[a] = idx ++ ;
}
int bfs()
{
int hh = 0, tt = 0;
q[0] = 1;
memset(d, -1, sizeof d);
d[1] = 0;
while(hh <= tt)
{
int t = q[hh ++];
for(int i = h[t]; i != -1; i = ne[i])
{
int j = e[i];
if(d[j] == -1)
{
d[j] = d[t] + 1;
q[ ++ tt] = j;
}
}
}
return d[n];
}
int main()
{
cin >> n >> m;
memset(h, -1, sizeof h);
for(int i = 0; i < m; i ++)
{
int a, b;
cin >> a >> b;
add(a, b);
}
cout << bfs() << endl;
return 0;
}
标签:遍历,idx,int,memset,ne,include,宽度,sizeof,例题
From: https://blog.51cto.com/u_16492348/9644156