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Roadblocks

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 14531 Accepted: 5107

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 length D (1 ≤ D ≤ 5000)

Output

Line 1: The length of the second shortest path between node 1 and node N

Sample Input

4 41 2 1002 4 2002 3 2503 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)

Source

USACO 2006 November Gold

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题目大意：求起点到终点的次短路长度。

思路：dijkstra算法。数组同时保存最短路和次短路，判断时除去判断是否路径权值小于当前最短路外还要判断是否大于最短路且小于当前次短路。

附上AC代码：

#include<iostream>#include<queue>#include<algorithm>#include<vector>using namespace std;const int maxn = 5000 + 5;const int Inf = 0x7f7f7f;int dis[maxn];//最短路径int dis2[maxn];//次短路径typedef pair<int, int>P;//first=最短路，second=路口标号int n, r;struct edge {int to, dis;};vector<edge>G[maxn];void dijkstra(){priority_queue<P, vector<P>, greater<P> >pq;fill(dis + 1, dis + 1 + n, Inf);fill(dis2 + 1, dis2 + 1 + n, Inf);dis[1] = 0;pq.push(P(0, 1));while (!pq.empty()){P p = pq.top();pq.pop();int v = p.second, d = p.first;if (dis2[v] < d)continue;//当前次短路径小于标记的最短路径，已经扫描过for (int i = 0; i < G[v].size(); i++){edge e = G[v][i];int d2 = d + e.dis;if (dis[e.to] > d2) {swap(dis[e.to], d2);pq.push(P(dis[e.to], e.to));}if (dis[e.to] < d2 && dis2[e.to] > d2) {dis2[e.to] = d2;pq.push(P(dis2[e.to], e.to));}}}cout << dis2[n] << endl;}int main(){ios::sync_with_stdio(false);cin >> n >> r;for (int i = 0; i < r; i++){edge e;int from;cin >> from >> e.to >> e.dis;G[from].push_back(e);swap(from, e.to);G[from].push_back(e);}dijkstra();//system("pause");return 0;}