BFS与最短路问题小结 - AcWing

题目 这题就可以BFS（边权值相同）（g[i][j]也可以为inf，g[i][i]还能为0）

代码

#include <bits/stdc++.h>
using namespace std;
const int N = 510;
const int inf = 0x3f3f3f3f;
int n, m;
int a[N];
int dist[N];
int g[N][N];
bool st[N];
int dijkstra()  // 求1号点到n号点的最短路距离
{
    memset(dist, 0x3f, sizeof dist);
    dist[1] = 0;
    
    for(int i = 0; i < n; i++)
    {
        int t = -1;
        for(int j = 1; j <= n; j++)
        {
            if(!st[j] && (t == -1 || dist[j] < dist[t]))
                t = j;
        }
        
        st[t] = 1;
        for(int j = 1; j <= n; j++)
            dist[j] = min(dist[j], dist[t] + g[t][j]);
    }
    
    return dist[n] - 1;
}

int main()
{
    cin >> m >> n;
    memset(g, 0x3f, sizeof g);
    for(int i = 1; i <= n; i++)
        g[i][i] = 0;
        
    getchar();
    for(int i = 1; i <= m; i++)
    {
        string line;
        getline(cin, line);  // 读取整行输入
        stringstream ss(line);
        int station, cnt = 0;
        while (ss >> station)  // 逐个读取每一站的编号（成功返回引用，失败返回nullptr）, 等同于scanf（成功返回成功个数，失败返回0或负）
            a[++cnt] = station;
        
        for (int j = 1; j <= cnt; j++) {
            for (int k = j + 1; k <= cnt; k++) {
                g[a[j]][a[k]] = min(g[a[j]][a[k]], 1);
            }
        }
    }
    
    int t = dijkstra();
    cout << (t > inf/2 ? "NO" : to_string(t));
}