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。
1 #include <map>
2 #include <set>
3 #include <list>
4 #include <queue>
5 #include <stack>
6 #include <cmath>
7 #include <vector>
8 #include <cstdio>
9 #include <string>
10 #include <cstring>
11 #include <cstdlib>
12 #include <iostream>
13 #include <algorithm>
14 using namespace std;
15 typedef long long ll;
16
17 #define eps 1e-8
18 #define inf 0x7f3f3f3f
19 #define debug puts("BUG")
20 #define read freopen("in.txt","r",stdin)
21
22 #define N 111
23 #define M 11111
24 struct node
25 {
26 int v,w,n;
27 }ed[M],ed2[M];
28 int head[N],head2[N],cnt,cnt2;
29 struct str
30 {
31 int v,p;
32 }st[M*10];
33 struct node2
34 {
35 int v,g,f,dx;
36 ll m1,m2;
37 void set(int _v,int _g,int _f,int dx,ll m1,ll m2)
38 {
39 this->v = _v;
40 this->g = _g;
41 this->f = _f;
42 this->dx = dx;
43 this->m1 = m1;
44 this->m2 = m2;
45 }
46 bool operator < (const node2 &n) const
47 {
48 if (n.f == f) return n.g < g;
49 return n.f < f;
50 }
51 };
52 void sc(int s,ll &m1,ll &m2)
53 {
54 if (s>50) m2 |= ((s-51)<<1);
55 else m1 |= ((s-1)<<1);
56 }
57 bool ck(int s,ll m1,ll m2)
58 {
59 if (s>50) return (m2&((s-51)<<1)) != 0;
60 else return (m1&((s-1)<<1)) != 0;
61 }
62 void init()
63 {
64 memset(head,-1,sizeof(head));
65 memset(head2,-1,sizeof(head2));
66 cnt = cnt2 = 0;
67 }
68 void add(int u,int v,int w)
69 {
70 ed[cnt].v = v;
71 ed[cnt].w = w;
72 ed[cnt].n = head[u];
73 head[u] = cnt++;
74 ed2[cnt2].v = u;
75 ed2[cnt2].w = w;
76 ed2[cnt2].n = head2[v];
77 head2[v] = cnt2++;
78 }
79 int h[N];
80 bool vis[N];
81 void spfa(int s,int n)
82 {
83 queue<int>q;
84 memset(vis,0,sizeof(vis));
85 for (int i = 1; i <= n; ++i)
86 h[i] = inf;
87 q.push(s);
88 h[s] = 0;
89 vis[s] = true;
90 while (!q.empty())
91 {
92 int u = q.front();
93 q.pop();
94 vis[u] = false;
95 for (int i=head2[u];~i;i=ed2[i].n)
96 {
97 int v = ed2[i].v,w = ed2[i].w;
98 if (h[v]>h[u]+w)
99 {
100 h[v] = h[u] + w;
101 if (!vis[v])
102 {
103 vis[v] = true;
104 q.push(v);
105 }
106 }
107 }
108 }
109 }
110 int ans[N];
111 void astar(int s,int t,int k)
112 {
113 priority_queue<node2> q;
114 int cnt = 0, dx = 0;
115 node2 n,n2;
116 ll m1=0,m2=0;
117 sc(s,m1,m2);
118 st[dx].p = -1,st[dx].v = s;
119 n.set(s,0,h[s],dx++,m1,m2);
120 q.push(n);
121 while (!q.empty())
122 {
123 n = q.top();
124 q.pop();
125 if (n.v == t) cnt++;
126 if (cnt == k)
127 {
128 int a = n.dx, b = 0;
129 while (~a)
130 {
131 ans[b++] = st[a].v;
132 a = st[a].p;
133 }
134 printf("%d %d\n",n.g,b);
135 bool ff = false;
136 for (int i = b-1; i >= 0; --i)
137 {
138 if (ff) printf(" ");
139 else ff = true;
140 printf("%d",ans[i]);
141 }puts("");
142 return ;
143 }
144 for (int i=head[n.v];~i;i=ed[i].n)
145 {
146 int v = ed[i].v, w = ed[i].w;
147 m1 = n.m1, m2 = n.m2;
148 if (ck(v,m1,m2)) continue;
149 sc(v,m1,m2);
150 st[dx].p = n.dx, st[dx].v = v;
151 n2.set(v,n.g+w,n.g+w+h[v],dx++,m1,m2);
152 q.push(n2);
153 }
154 }
155 return ;
156 }
157 int main()
158 {
159 //read;
160 int n,m,k;
161 while (~scanf("%d%d%d",&n,&m,&k))
162 {
163 init();
164 int u,v,w,s,t;
165 for (int i = 0; i < m; ++i)
166 {
167 scanf("%d%d%d",&u,&v,&w);
168 add(u,v,w);
169 }
170 scanf("%d%d",&s,&t);
171 spfa(t,n);
172 astar(s,t,k);
173 }
174 return 0;
175 }
时间： 2024-12-20 01:20:22