晚上匆忙整理了下，有时间继续！加油！

Dijkstra

void Dijkstra(){
    priority_queue<pair<int,int>>q;
    memset(d,0x3f,sizeof d);
    memset(v,0,sizeof v);
    d[1] = 0;
    q.push({0,1});               //利用小根堆
    while(q.size()) {
        int x = q.top().second;
        q.pop();
        if(v[x]) continue;
        v[x] = 1;
        for(int i = head[x]; i ; i = Next[i]) {
            int y = ver[i];
            int z = edge[i];
            if(d[y] > d[x] + z) {
                d[y] = d[x] + z;
                q.push({-d[y],y});
            }
        }
    }
}

SPFA

void SPFA() {
    queue<int>q;
    memset(d,0x3f,sizeof d);
    memset(v,0,sizeof v);
    d[1] = 0;
    v[1] = 1;
    q.push(1);               
    while(q.size()) {
        int x = q.front();
        q.pop();
        v[x] = 0;
        for(int i = head[x]; i ; i = Next[i]) 
            int y = ver[i];
            int z = edge[i];
            if(d[y] > d[x] + z) 
                d[y] = d[x] + z;
                if(!v[y])q.push(y),v[y] = 1;
            }
        }
    }
}

Floyd

for(int k = 1; k <= n; k ++ )
    for(int i = 1; i <= n; i ++ )
        for(int j = 1; j <= n; j ++ )
            d[i][j] = min(d[i][j],d[i][k]+d[k][j]);

好简洁啊

传递闭包

for(int k = 1; k <= n; k ++ )
   for(int i = 1; i <= n; i ++ )
      for(int j = 1; j <= n; j ++ )
          d[i][j] | = d[i][k] & d[k][j];