第K短路

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Dijkstra

/* * Dijkstra变形，可以证明每个点经过的次数为小于等于K， * 所有Dijkstra的数组dist由一维变为二维，记录经过该点 * 1次、2次......k次的最小值 * 输出dist[n - 1][k]即可 */int g[1010][1010];int n, m, x;const int INF = 0x3f3f3f3f;int vis[1010];int dist[1010][20];int main(int argc, const char * argv[]){    while (cin >> n >> m >> x)    {        //初始化        memset(g, 0x3f, sizeof(g));        memset(dist, 0x3f, sizeof(dist));        memset(vis, 0, sizeof(vis));        for (int i = 0; i < m; i++)        {            int p, q, r;            cin >> p >> q >> r;            if (r < g[p][q])            {                g[p][q] = r;            }        }        dist[1][0] = 0;        dist[0][0] = INF;        while (1)        {            int k = 0;            for (int i = 1; i <= n; i++)            {                if (vis[i] < x && dist[i][vis[i]] < dist[k][0])                {                    k = i;                }            }            if (k == 0)            {                break;            }            if (k == n && vis[n] == x - 1)            {                break;            }            for (int i = 1; i <= n; i++)            {                if (vis[i] < x && dist[k][vis[k]] + g[k][i] < dist[i][x])                {                    dist[i][x] = dist[k][vis[k]] + g[k][i];                    for (int j = x; j > 0; j--)                    {                        if (dist[i][j] < dist[i][j - 1])                        {                            swap(dist[i][j], dist[i][j - 1]);                        }                    }                }            }            vis[k]++;        }        if (dist[n][x - 1] < INF)        {            cout << dist[n][x - 1] << endl;        }        else        {            cout << -1 << endl;        }    }    return 0;}

A*

/* * A* 估价函数  fi为到当前点走过的路经长度，hi为该点到终点的长度 * gi = hi + fi */int n, m, x, ct;int g[1010][1010];int gr[1010][1010];int dist[1010];int vis[1010];const int INF = 0x3f3f3f3f;struct node{    int id;    int fi;    int gi;    friend bool operator < (node a, node b)    {        if (a.gi == b.gi)        {            return a.fi > b.fi;        }        return a.gi > b.gi;    }} s[20000010];int init(){    memset(dist, 0x3f, sizeof(dist));    for (int i = 0; i <= n; i++)    {        vis[i] = 1;    }    dist[n - 1] = 0;    for (int i = 0; i < n; i++)    {        int k = n;        for (int j = 0; j < n; j++)        {            if (vis[j] && dist[j] < dist[k])            {                k = j;            }        }        if (k == n)        {            break;        }        vis[k] = 0;        for (int j = 0; j < n; j++)        {            if (vis[j] && dist[k] + gr[k][j] < dist[j])            {                dist[j] = dist[k] + gr[k][j];            }        }    }    return 1;}int solve(){    if (dist[0] == INF)    {        return -1;    }    ct = 0;    s[ct].id = 0;    s[ct].fi = 0;    s[ct++].gi = dist[0];    int cnt = 0;    while (ct)    {        int id = s[0].id;        int fi = s[0].fi;        if (id == n - 1)        {            cnt++;        }        if (cnt == x)        {            return fi;        }        pop_heap(s, s + ct);        ct--;        for (int j = 0; j < n; j++)        {            if (g[id][j] < INF)            {                s[ct].id = j;                s[ct].fi = fi + g[id][j];                s[ct].gi = s[ct].fi + dist[j];                ct++;                push_heap(s, s + ct);            }        }    }    return -1;}int main(){    while (cin >> n >> m >> x)    {        memset(g, 0x3f, sizeof(g));        memset(gr, 0x3f, sizeof(gr));        for (int i = 0; i < n; i++)        {            int p, q, r;            cin >> p >> q >> r;            p--;            q--;            g[p][q] = g[p][q] <= r ? g[p][q] : r;            gr[q][p] = gr[q][p] <= r ? gr[q][p] : r;        }        init();        cout << solve() << endl;    }    return 0;}

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