POJ-3255 Roadblocks---Dijkstra队列优化+次短路

Posted 2020-10-30 努力努力再努力x

题目链接：

https://vjudge.net/problem/POJ-3255

题目大意：

给无向图，求1到n的次短路长度

思路：

由于边数较多，应该使用dijkstra的队列优化

用d数组存储最短路，用d2数组存储次短路，每次更新的时候，先松弛更新最短路，如果松弛更新成功，把之前的最短路取出，再和次短路比较，更新次短路。每次更新两个数组

 1 #include<iostream>
 2 #include<vector>
 3 #include<queue>
 4 #include<algorithm>
 5 #include<cstring>
 6 #include<cstdio>
 7 using namespace std;
 8 typedef pair<int, int> Pair;
 9 const int maxn = 5000 + 10;
10 const int INF = 0x3f3f3f3f;
11 int n, m;
12 struct edge
13 {
14     int v, w;
15     edge(int v, int w):v(v), w(w){}
16 };
17 vector<edge>G[maxn];
18 int d[maxn], d2[maxn];
19 void dijkstra()
20 {
21     memset(d, INF, sizeof(d));
22     memset(d2, INF, sizeof(d2));
23     d[1] = 0;
24     priority_queue<Pair, vector<Pair>, greater<Pair> >q;
25     q.push(Pair(0, 1));//1为当前节点，0位最短距离
26     while(!q.empty())
27     {
28         Pair now = q.top();
29         q.pop();
30         int dis = now.first, u = now.second;
31         //cout<<dis<<" "<<u<<endl;
32         if(dis > d2[u])continue;//距离比次短距离大，直接continue
33         for(int i = 0; i < G[u].size(); i++)
34         {
35             edge &e = G[u][i];
36             int dis2 = dis + e.w;
37             if(d[e.v] > dis2)
38             {
39                 swap(d[e.v], dis2);//dis2保存较小的值和后面进行比较
40                 q.push(Pair(d[e.v], e.v));
41             }
42             if(d2[e.v] > dis2 && d[e.v] < dis2)//保证dis2为次短路
43             {
44                 d2[e.v] = dis2;
45                 q.push(Pair(d2[e.v], e.v));//次短路也入队列
46             }
47         }
48     }
49     cout<<d2[n]<<endl;
50 }
51 int main()
52 {
53     cin >> n >> m;
54     int u, v, w;
55     while(m--)
56     {
57         scanf("%d%d%d", &u, &v, &w);
58         G[u].push_back(edge(v, w));
59         G[v].push_back(edge(u, w));
60     }
61     dijkstra();
62 }