洛谷

https://www.luogu.org/problem/P1346
使用最短路之前居然忘记清空了。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int MAXN = 1005;

int n, s, t;
int dis[MAXN];
bool vis[MAXN];
int G[MAXN][MAXN];

const int INF = 0x3f3f3f3f;

priority_queue<pair<int, int> >pq;
void Dijkstra(int s) {
    for(int i=1;i<=n;++i)
        dis[i]=INF;
    dis[s] = 0;
    pq.push({-dis[s], s});
    while(!pq.empty()) {
        int u = pq.top().second;
        pq.pop();
        if(vis[u])
            continue;
        /*printf("u=%d
",u);
        for(int i = 1; i <= n; ++i) {
            printf(" %d", i, dis[i]);
        }
        puts("");*/
        vis[u] = 1;
        for(int v = 1; v <= n; ++v) {
            int w = G[u][v];
            if(!vis[v] && dis[u] + w < dis[v]) {
                dis[v] = dis[u] + w;
                pq.push({-dis[v], v});
            }
        }
    }
}

int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
#endif // Yinku
    scanf("%d%d%d", &n, &s, &t);
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= n; ++j)
            G[i][j] = INF;
    for(int i = 1; i <= n; ++i)
        G[i][i] = 0;
    for(int u = 1; u <= n; ++u) {
        int k;
        scanf("%d", &k);
        for(int j = 1; j <= k; ++j) {
            int v;
            scanf("%d", &v);
            if(j == 1)
                G[u][v] = 0;
            else
                G[u][v] = 1;
        }
    }
    Dijkstra(s);
    /*for(int i = 1; i <= n; ++i) {
        printf("%d: %d
", i, dis[i]);
    }*/
    printf("%d
", dis[t] < INF ? dis[t] : -1);
    return 0;
}

看了一下题解貌似还有更快的。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int MAXN = 1005;

int n, s, t;
int dis[MAXN];
bool vis[MAXN];
int G[MAXN][MAXN];

const int INF = 0x3f3f3f3f;

deque<int>q;
void BFS(int s) {
    for(int i = 1; i <= n; ++i)
        dis[i] = INF;
    dis[s] = 0;
    q.push_front(s);
    while(!q.empty()) {
        int u = q.front();
        q.pop_front();
        if(vis[u])
            continue;
        vis[u] = 1;
        for(int v = 1; v <= n; ++v) {
            if(vis[v])
                continue;
            int w = G[u][v];
            if(w == 0) {
                if(dis[u] < dis[v]) {
                    dis[v] = dis[u];
                    q.push_front(v);
                }
            } else if(w == 1) {
                if(dis[u] + 1 < dis[v]) {
                    dis[v] = dis[u] + 1;
                    q.push_back(v);
                }
            }
        }
    }
}

int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
#endif // Yinku
    scanf("%d%d%d", &n, &s, &t);
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= n; ++j)
            G[i][j] = INF;
    for(int i = 1; i <= n; ++i)
        G[i][i] = 0;
    for(int u = 1; u <= n; ++u) {
        int k;
        scanf("%d", &k);
        for(int j = 1; j <= k; ++j) {
            int v;
            scanf("%d", &v);
            G[u][v] = (j != 1);
        }
    }
    BFS(s);
    printf("%d
", dis[t] < INF ? dis[t] : -1);
    return 0;
}

那应该可以得到一个使得Dijkstra更快的办法，就是另外开一个队列，把u节点相连的距离为0的节点加入队列，每次优先从队列里面取，其次才从优先队列里面取。

这个01BFS是指有花费的边的权都一样，所以不需要优先队列，从u出发的能到的点假如有花费一定是出现在队尾的。