Roadblocks
Time Limit: 2000ms
Memory Limit: 65536KB
This problem will be judged on PKU. Original ID: 3255
64-bit integer IO format: %lld Java class name: Main
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 length D (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
Source
解题：次短路模板题。。。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <cmath>
5 #include <algorithm>
6 #include <climits>
7 #include <vector>
8 #include <queue>
9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #include <stack>
13 #define LL long long
14 #define pii pair<long long,int>
15 #define INF 0x3f3f3f3f
16 using namespace std;
17 struct arc{
18 int to,w;
19 arc(int x = 0,int y = 0){
20 to = x;
21 w = y;
22 }
23 };
24 const int maxn = 5010;
25 vector<arc>g[maxn];
26 int d[maxn],d2[maxn],n,m;
27 priority_queue< pii,vector< pii >,greater< pii > >q;
28 void dijkstra(){
29 for(int i = 0; i <= n; i++) d[i] = d2[i] = INF;
30 while(!q.empty()) q.pop();
31 d[1] = 0;
32 q.push(make_pair(d[1],1));
33 while(!q.empty()){
34 int u = q.top().second;
35 int w = q.top().first;
36 q.pop();
37 if(d2[u] < w) continue;
38 for(int i = 0; i < g[u].size(); i++){
39 int dis = w + g[u][i].w;
40 if(d[g[u][i].to] > dis){
41 swap(dis,d[g[u][i].to]);
42 q.push(make_pair(d[g[u][i].to],g[u][i].to));
43 }
44 if(d2[g[u][i].to] > dis && d[g[u][i].to] < dis){
45 d2[g[u][i].to] = dis;
46 q.push(make_pair(d2[g[u][i].to],g[u][i].to));
47 }
48 }
49 }
50 }
51 int main(){
52 int u,v,w;
53 while(~scanf("%d %d",&n,&m)){
54 for(int i = 0; i <= n; i++) g[i].clear();
55 for(int i = 0; i < m; i++){
56 scanf("%d %d %d",&u,&v,&w);
57 g[u].push_back(arc(v,w));
58 g[v].push_back(arc(u,w));
59 }
60 dijkstra();
61 printf("%d\n",d2[n]);
62 }
63 return 0;
64 }