本文地址：https://www.cnblogs.com/maplefighting/p/9124073.html
题目名称：Roadblocks
链接：http://poj.org/problem?id=3255
题意：有 N 个点，R条边，计算从 1 出发到 N的次短路长度，次短路是指长度比那些最短路长的路径。
思路：emmm，最短路是每次保存最短的点，那我们每次保存最短和次短就好了
代码如下：
1 #include<cstdio>
2 #include<iostream>
3 #include<queue>
4 using namespace std;
5 typedef pair<int,int> P;
6 const int INF = 0x3f3f3f3f;
7 const int maxn = 5005;
8 struct Edge{
9 int v, w;
10 int nxt;
11 }edge[100005 * 2];
12 int head[maxn];
13 int n, r;
14 int cnt, dist1[maxn], dist2[maxn];
15 void add_edge(int u, int v, int w){
16 edge[cnt].v = v;
17 edge[cnt].w = w;
18 edge[cnt].nxt = head[u];
19 head[u] = cnt++;
20 }
21 void init(){
22 cnt = 0;
23 fill(head, head + n + 1, -1);
24 fill(dist1, dist1 + n + 1, INF);
25 fill(dist2, dist2 + n + 1, INF);
26 }
27 void spfa(int s){
28 priority_queue<P, vector<P>, greater<P> > Q;
29 dist1[s] = 0;
30 Q.push(P(0, 1));
31 while(!Q.empty()){
32 P p = Q.top(); Q.pop();
33 int v = p.second, w = p.first;
34 if(dist2[v] < w) continue;
35 for(int i = head[v]; i != -1; i = edge[i].nxt){
36 Edge &e = edge[i];
37 int d = w + e.w;
38 if(dist1[e.v] > d){
39 swap(dist1[e.v], d);
40 Q.push(P(dist1[e.v], e.v));
41 }
42 if(dist2[e.v] > d && dist1[e.v] < d){
43 dist2[e.v] = d;
44 Q.push(P(dist2[e.v], e.v));
45 }
46 }
47 }
48 }
49 int main(){
50
51 scanf("%d%d", &n, &r);
52 init();
53 for(int i = 1; i <= r; i++){
54 int a,b,d;
55 scanf("%d%d%d", &a, &b, &d);
56 add_edge(a, b, d);
57 add_edge(b, a, d);
58 }
59 spfa(1);
60 printf("%d\n", dist2[n]);
61 return 0;
62 }