最短路径模板

spfa:

void add(int u,int v,int w)
{
    g[cnt].v = v;
    g[cnt].w = w;
    g[cnt].next = pre[u];
    pre[u] = cnt++;
}
bool spfa(int s)
{
    int i;
    queue<int>q;
    for (i = 0; i < maxn; ++i) dis[i] = -inf;
    dis[s] = 0;
    q.push(s); inq[s] = true;
    while (!q.empty())
    {
        int u = q.front(); q.pop();
        if (++ind[u] > 24) return false;
        inq[u] = false;
        for (i = pre[u]; i != -1; i = g[i].next)
        {
            int v = g[i].v, w = g[i].w;
            if (dis[v] < dis[u] + w)
            {
                dis[v] = dis[u] + w;
                if (!inq[v])
                {
                    inq[v] = true;
                    q.push(v);
                }
            }
        }
    }
    return true;
}

  

 

Dijkstra

void Dijkstra(int n, int v, int *dist, int *prev, int c[maxnum][maxnum])
{
    bool s[maxnum];    // 判断是否已存入该点到S集合中
    for(int i=1; i<=n; ++i)
    {
        dist[i] = c[v][i];
        s[i] = 0;     // 初始都未用过该点
        if(dist[i] == maxint)
            prev[i] = 0;
        else
            prev[i] = v;
    }
    dist[v] = 0;
    s[v] = 1;
  
    // 依次将未放入S集合的结点中,取dist[]最小值的结点,放入结合S中
    // 一旦S包含了所有V中顶点,dist就记录了从源点到所有其他顶点之间的最短路径长度
         // 注意是从第二个节点开始,第一个为源点
    for(int i=2; i<=n; ++i)
    {
        int tmp = maxint;
        int u = v;
        // 找出当前未使用的点j的dist[j]最小值
        for(int j=1; j<=n; ++j)
            if((!s[j]) && dist[j]<tmp)
            {
                u = j;              // u保存当前邻接点中距离最小的点的号码
                tmp = dist[j];
            }
        s[u] = 1;    // 表示u点已存入S集合中
  
        // 更新dist
        for(int j=1; j<=n; ++j)
            if((!s[j]) && c[u][j]<maxint)
            {
                int newdist = dist[u] + c[u][j];
                if(newdist < dist[j])
                {
                    dist[j] = newdist;
                    prev[j] = u;
                }
            }
    }
}
  
// 查找从源点v到终点u的路径,并输出
void searchPath(int *prev,int v, int u)
{
    int que[maxnum];
    int tot = 1;
    que[tot] = u;
    tot++;
    int tmp = prev[u];
    while(tmp != v)
    {
        que[tot] = tmp;
        tot++;
        tmp = prev[tmp];
    }
    que[tot] = v;
    for(int i=tot; i>=1; --i)
        if(i != 1)
            cout << que[i] << " -> ";
        else
            cout << que[i] << endl;
}

  

 

bellman_ford

void relax(int u, int v, int weight)
{
    if(dist[v] > dist[u] + weight)
        dist[v] = dist[u] + weight;
}
  
bool Bellman_Ford()
{
    for(int i=1; i<=nodenum-1; ++i)
        for(int j=1; j<=edgenum; ++j)
            relax(edge[j].u, edge[j].v, edge[j].weight);
    bool flag = 1;
    // 判断是否有负环路
    for(int i=1; i<=edgenum; ++i)
        if(dist[edge[i].v] > dist[edge[i].u] + edge[i].weight)
        {
            flag = 0;
            break;
        }
    return flag;
}

 floyd

void floyd()
{
    int i,j,k;
    for (k = 0; k < n; ++k)
    {
        for (i = 0; i < n; ++i)
        {
            for (j = 0; j < n; ++j)
            {
                if (map[i][k] != inf && map[k][j] != inf
                    && map[i][k] + map[k][j] < map[i][j])
                map[i][j] = map[i][k]*map[k][j];
            }
        }
    }
}
posted @ 2012-08-18 00:40  山路水桥  阅读(230)  评论(0编辑  收藏  举报