单源最短路径

单源最短路径

Floyd

$O(N^3)$

int Floyd(int start, int end){
    memset(dis, Inf, sizeof(dis));
    for(int k = 1; k <= N; k++){
        for(int i = 1; i <= N; i++){
            for(int j = 1; j <= N; j++){
                dis[i][j] = min(dis[i][j], dis[i][k]+dis[k][j]);
            }
        }
    }
    return dis[start][end];
}

Dijkstra(heap)

$O(M+NlgN)$

int Dijkstra(int start, int end){
    memset(dis, Inf, sizeof(dis));
    memset(vis, 0, sizeof(vis));
    typedef pair<int, int> Pair;
    priority_queue< Pair, vector<Pair>, greater<Pair> > que;
    que.push(make_pair((dis[start]=0), start));
    while(!que.empty()){
        Pair cnt = que.top();
        que.pop();
        int now = cnt.second;
        if(vis[now])  continue;
        vis[now] = 1;
        int t = st[now];
        while(t != 0){
            if(dis[H[t].v] > dis[now]+H[t].w){
                dis[H[t].v] = dis[now] + H[t].w;
                if(!vis[H[t].v])  que.push(make_pair(dis[H[t].v], H[t].v));
            }
            t = H[t].next;
        }
    }
    return dis[end];
}

Spfa

$O(MN)$

int Spfa(int start, int end){
    memset(dis, Inf, sizeof(dis));
    memset(vis, 0, sizeof(vis));
    queue<int> que;
    que.push(start);
    vis[start] = 1; 
    dis[start] = 0;
    while(!que.empty()){
        int now = que.front();
        que.pop();
        int t = st[now];
        vis[now] = 0;
        while(t != 0){
            if(dis[H[t].v] > dis[now]+H[t].w){
                dis[H[t].v] = dis[now] + H[t].w;
                if(!vis[H[t].v])  que.push(H[t].v);
                vis[H[t].v] = 1;
            }
            t = H[t].next;
        }
    }
    return dis[end];
}

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注意：
memset(,,sizeof()) 记得写头文件 #include<cstring>
从网上抠下来一张表：

	稠密图	稀疏图	有负权边
单源问题	Dijkstra+heap	SPFA (或Dijkstra+heap,根据稀疏程度)	SPFA
APSP(无向图)	SPFA(优化)/Floyd	SPFA(优化)	SPFA(优化)
APSP(有向图)	Floyd	SPFA (或Dijkstra+heap,根据稀疏程度)	SPFA

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但是某次考试 单源 稠密图 Dijkstra+heap 迷之比 Spfa 慢 3~4倍……