最短路径模板

spfa:

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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

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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

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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

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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];

            }

        }

    }

}