SGU 145

节点不可重复经过的K短路问题。

思路：二分路径长度，深搜小于等于路径长度的路径数。可以利用可重复点K短路问题中的A*函数进行剪枝。

尝试另一种解法：把可重复点K短路A*直接搬过来，堆中的每个元素额外记录之前走过的所有点。这样就可以据此防止走重复的点。最大100个点，可用两个长整形状态压缩。

一直PE,无法验证效率。

 #include <map>
 #include <set>
 #include <list>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <vector>
 #include <cstdio>
 #include <string>
 #include <cstring>
 #include <cstdlib>
 #include <iostream>
 #include <algorithm>
 using namespace std;
 typedef long long  ll;

 #define eps 1e-8
 #define inf 0x7f3f3f3f
 #define debug puts("BUG")
 #define read freopen("in.txt","r",stdin)

 #define N 111
 #define M 11111
 struct node
 {
     int v,w,n;
 }ed[M],ed2[M];
 int head[N],head2[N],cnt,cnt2;
 struct str
 {
     int v,p;
 }st[M*];
 struct node2
 {
     int v,g,f,dx;
     ll m1,m2;
     void set(int _v,int _g,int _f,int dx,ll m1,ll m2)
     {
         this->v = _v;
         this->g = _g;
         this->f = _f;
         this->dx = dx;
         this->m1 = m1;
         this->m2 = m2;
     }
     bool operator < (const node2 &n) const
     {
         if (n.f == f) return n.g < g;
         return n.f < f;
     }
 };
 void sc(int s,ll &m1,ll &m2)
 {
     if (s>) m2 |= ((s-)<<);
     else m1 |= ((s-)<<);
 }
 bool ck(int s,ll m1,ll m2)
 {
     if (s>) return (m2&((s-)<<)) != ;
     else return (m1&((s-)<<)) != ;
 }
 void init()
 {
     memset(head,-,sizeof(head));
     memset(head2,-,sizeof(head2));
     cnt = cnt2 = ;
 }
 void add(int u,int v,int w)
 {
     ed[cnt].v = v;
     ed[cnt].w = w;
     ed[cnt].n = head[u];
     head[u] = cnt++;
     ed2[cnt2].v = u;
     ed2[cnt2].w = w;
     ed2[cnt2].n = head2[v];
     head2[v] = cnt2++;
 }
 int h[N];
 bool vis[N];
 void spfa(int s,int n)
 {
     queue<int>q;
     memset(vis,,sizeof(vis));
     for (int i = ; i <= n; ++i)
         h[i] = inf;
     q.push(s);
     h[s] = ;
     vis[s] = true;
     while (!q.empty())
     {
         int u = q.front();
         q.pop();
         vis[u] = false;
         for (int i=head2[u];~i;i=ed2[i].n)
         {
             int v = ed2[i].v,w = ed2[i].w;
             if (h[v]>h[u]+w)
             {
                 h[v] = h[u] + w;
                 if (!vis[v])
                 {
                     vis[v] = true;
                     q.push(v);
                 }
             }
         }
     }
 }
 int ans[N];
 void astar(int s,int t,int k)
 {
     priority_queue<node2> q;
     int cnt = , dx = ;
     node2 n,n2;
     ll m1=,m2=;
     sc(s,m1,m2);
     st[dx].p = -,st[dx].v = s;
     n.set(s,,h[s],dx++,m1,m2);
     q.push(n);
     while (!q.empty())
     {
         n = q.top();
         q.pop();
         if (n.v == t) cnt++;
         if (cnt == k)
         {
             int a = n.dx, b = ;
             while (~a)
             {
                 ans[b++] = st[a].v;
                 a = st[a].p;
             }
             printf("%d %d\n",n.g,b);
             bool ff = false;
             for (int i = b-; i >= ; --i)
             {
                 if (ff) printf(" ");
                 else ff = true;
                 printf("%d",ans[i]);
             }puts("");
             return ;
         }
         for (int i=head[n.v];~i;i=ed[i].n)
         {
             int v = ed[i].v, w = ed[i].w;
             m1 = n.m1, m2 = n.m2;
             if (ck(v,m1,m2)) continue;
             sc(v,m1,m2);
             st[dx].p = n.dx, st[dx].v = v;
             n2.set(v,n.g+w,n.g+w+h[v],dx++,m1,m2);
             q.push(n2);
         }
     }
     return ;
 }
 int main()
 {
     //read;
     int n,m,k;
     while (~scanf("%d%d%d",&n,&m,&k))
     {
         init();
         int u,v,w,s,t;
         for (int i = ; i < m; ++i)
         {
             scanf("%d%d%d",&u,&v,&w);
             add(u,v,w);
         }
         scanf("%d%d",&s,&t);
         spfa(t,n);
         astar(s,t,k);
     }
     return ;
 }