给出1～N 个点 的距离， 求从1号到N号的次短路， 直接用k短路来做了，，dj会TLE， 用spfa就过了
题目：
I - Roadblocks
Time Limit:2000MS Memory
Limit:65536KB 64bit IO
Format:%I64d & %I64u
Submit Status
Description
Bessie has moved to a small farm and sometimes enjoys returning to visit one
of her best friends. She does not want to get to her old home too quickly,
because she likes the scenery along the way. She has decided to take the
second-shortest rather than the shortest path. She knows there must be some
second-shortest path.
The countryside consists of R (1 ≤ R ≤
100,000) bidirectional roads, each linking two of the N (1
≤ N ≤ 5000) intersections, conveniently numbered
1..N. Bessie starts at intersection 1, and her friend (the destination)
is at intersection N.
The second-shortest path may share roads with any of the shortest paths, and
it may backtrack i.e., use the same road or intersection more than once. The
second-shortest path is the shortest path whose length is longer than the
shortest path(s) (i.e., if two or more shortest paths exist, the second-shortest
path is the one whose length is longer than those but no longer than any other
path).
Input
Line 1: Two space-separated
integers: N and R
Lines
2..R+1: Each line contains three space-separated
integers: A, B, and D that
describe a road that connects
intersections A and B and has
lengthD (1 ≤ D ≤ 5000)
Output
Line 1: The length of the second shortest path
between node 1 and node N
Sample Input
4 4
1 2 100
2 4 200
2 3 250
3 4 100
Sample Output
450
Hint
Two routes: 1 -> 2 -> 4 (length 100+200=300)
and 1 -> 2 -> 3 -> 4 (length 100+250+100=450)
代码：
1 #include <iostream>
2 #include <algorithm>
3 #include <vector>
4 #include <queue>
5 using namespace std;
6
7 #define MAXN 5000+10
8
9
10 int N,R;
11
12 #define MP(a,b) make_pair(a,b)
13 #define INF 100000000
14
15 #define pii pair<int,int>
16 vector<pii> G[MAXN];
17 struct pri
18 {
19 bool operator () (const pair<int,int> &p1,const pair<int,int>&p2)
20 {
21 return p1.second< p2.second;
22 }
23 };
24
25 int d[MAXN];
26 void dj(int s)
27 {
28 for(int i = 1;i<=N;i++)
29 {
30 d[i]= INF;
31 d[i] = INF;
32 }
33 d[s]=0;
34 d[s]=0;
35
36 priority_queue<pii,vector<pii>,pri> q;
37 q.push( MP(s,d[s]) );
38 while(!q.empty())
39 {
40 int u = q.top().first;
41 q.pop();
42
43 int size = G[u].size();
44
45 for(int i=0;i<size;i++)
46 {
47 int v,w;
48 v = G[u][i].first;
49 w = G[u][i].second;
50
51 if( d[v] > d[u]+w )
52 {
53 d[v]=d[u]+w;
54 q.push(MP(v,d[v]));
55
56 }
57 }
58 }
59
60 }
61 /*
62 void spfa(int src)
63 {
64
65 for(int i=1;i<=N;i++)
66 {
67 d[i] = INF;
68 }
69 d[src]=0;
70 queue<int> q;
71 q.push(src);
72 int a,next;
73 int vis[MAXN]={0};
74 while(!q.empty())
75 {
76 a = q.front();
77 q.pop();
78 vis[a] = 0;
79 int len = G[a].size();
80 for(int i=0;i<len;i++)
81 {
82 next = G[a][i].first;
83 if( d[next]> d[a]+G[a][i].second )
84 {
85 d[next] = d[a]+G[a][i].second;
86 if( !vis[next] )
87 {
88 q.push(next);
89 vis[next]=1;
90 }
91 }
92 }
93 }
94 }
95 */
96 struct node{
97
98 int now,g,h,f;
99 bool operator < (const node& x) const
100 {
101 if( f == x.f )
102 {
103 return x.g<g;
104 }
105 else
106 {
107 return x.f<f;
108 }
109 }
110 };
111 void astar(int src,int to)
112 {
113 priority_queue<node> q;
114 node a, next;
115
116 a.now = src;
117 a.g = 0;
118 a.f = d[src]+a.g;
119 q.push(a);
120 int cnt = 0;
121 while(!q.empty())
122 {
123 a = q.top();
124 q.pop();
125 if( a.now == to )
126 {
127 cnt++;
128 if( cnt == 2 )
129 {
130 cout<<a.g<<endl;
131 return;
132 }
133 }
134 int size = G[a.now].size();
135 for(int i=0;i<size;i++)
136 {
137 next.now = G[a.now][i].first;
138 next.g = a.g + G[a.now][i].second;
139 next.f = next.g+d[next.now];
140 q.push(next);
141 }
142 }
143 }
144
145 int main()
146 {
147
148 cin>>N>>R;
149
150 for(int i=0;i<R;i++)
151 {
152 int f,t,w;
153 cin>>f>>t>>w;
154
155 G[f].push_back( MP(t,w) );
156 G[t].push_back( MP(f,w) );
157 }
158
159 dj(N);
160 // spfa(N);
161 astar(1,N);
162 return 0;
163 }
POJ3255:Roadblocks(次短路 SPFA+A星),码迷,mamicode.com