先从节点N广搜一边,确定每个点到N的最短距离d[1...N],
然后按到N的距离把节点分成各层,从第d[1]层开始,对每一层先求出本层到下一层的最小代价(字典序),然后由这个最小代价
求出下一层哪一个节点是可以(通过最优路径)到达的,进行标记.重复操作,知道第d[N] = 0层.遍历过程中直接输出即可.
附源代码:
View Code
1 //对图建立层的结构,然后分别处理每层即可
2 #include <iostream>
3 #include <fstream>
4 #include <algorithm>
5 #include <climits>
6 #include <cstdlib>
7 #include <cmath>
8 #include <cstdio>
9 #include <cstring>
10
11
12
13
14 using namespace std;
15
16 const int INF = INT_MAX;
17
18 typedef pair<int, int> pii;
19
20 typedef struct edge
21 {
22 int t;
23 int e;
24 edge* next;
25 } edge;
26
27 typedef struct node
28 {
29 int t;
30 node* next;
31 } node;
32
33 node pool[400000];
34 bool vis[110000];
35 node* link[110000];
36 edge g[1000000];
37 edge* head[110000];
38 int cnt;
39 int q[110000];
40
41 int N, M;
42 int d[110000];
43 int ans[110000];
44
45 inline void add_edge(int s, int t, int e)
46 {
47 g[cnt].t = t;
48 g[cnt].e = e;
49 g[cnt].next = head[s];
50 head[s] = &g[cnt++];
51
52 g[cnt].t = s;
53 g[cnt].e = e;
54 g[cnt].next = head[t];
55 head[t] = &g[cnt++];
56 }
57 inline void add_link(int s, int t)
58 {
59 pool[cnt].t = t;
60 pool[cnt].next = link[s];
61 link[s] = &pool[cnt++];
62 }
63
64 void build()
65 {
66 // memset(head, 0, sizeof(head));
67 // cnt = 0;
68
69 int s, t, e;
70 for(int i = 0; i < M; i++)
71 {
72 scanf("%d %d %d", &s, &t, &e);
73 add_edge(s, t, e);
74 }
75 }
76 void buildLink()
77 {
78 cnt = 0;
79 for(int i = 1; i <= N; i++)
80 add_link(d[1] - d[i], i);
81
82 }
83
84 int main()
85 {
86
87 // freopen("input.txt", "r", stdin);
88
89 scanf("%d %d", &N, &M);
90 build();
91 memset(d, -1, sizeof(d));
92
93 int s = 0, t = 0;;
94 d[N] = 0;
95 q[t++] = N;
96
97 while(s < t)
98 {
99 int u = q[s++];
100
101 for(edge* p = head[u]; p; p = p->next)
102 {
103 if(-1 == d[p->t])
104 {
105 d[p->t] = d[u] + 1;
106 q[t++] = p->t;
107 }
108 }
109 }
110 printf("%d\n", d[1]);
111 vis[1] = true;
112 buildLink();
113
114 for(int i = 0; i < d[1]; i++)
115 {
116 int ans = INF;
117 for(node* p = link[i]; p; p = p->next)
118 {
119 if(vis[p->t])
120 {
121 int x = p->t;
122 for(edge* tp = head[x]; tp; tp = tp->next)
123 {
124 if(d[x] - 1 == d[tp->t])
125 ans = min(ans, tp->e);
126 }
127 }
128 }
129 if(i)
130 putchar(' ');
131 printf("%d", ans);
132 if(i == d[1] - 1)
133 break;
134
135 for(node* p = link[i]; p; p = p->next)
136 {
137 if(vis[p->t])
138 {
139 int x = p->t;
140 for(edge* tp = head[x]; tp; tp = tp->next)
141 {
142 if(ans == tp->e && d[x] - 1 == d[tp->t])
143 {
144 vis[tp->t] = true;
145 }
146 }
147 }
148 }
149 }
150
151 puts("");
152
153 return 0;
154 }
~~~
