POJ 3255Roadblocks
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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
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
Sample Input
4 41 2 1002 4 2002 3 2503 4 100
Sample Output
450
Hint
Source
到某个顶点v的次短距离要么是到其他某个顶点u的最短距离再加上u到v的边,要么是到其他某个顶点的次短路再加上u到v的边。
采用和DIjkstra算法一样的思路,每次先维护最短距离,再维护次短距离。
#include <iostream>#include <queue>#include <vector>#include <algorithm>#include <utility>#include <cstdio>using namespace std;const int maxn =5000+10;#define INF 0x3f3f3f3ftypedef pair<int,int> P;int N,R;struct edge{ int to,cost;};vector<edge> G[maxn];int dist[maxn];//最短路int dist2[maxn];//次短路void solve(){ priority_queue<P,vector<P>,greater<P> > que; fill(dist,dist+N,INF); fill(dist2,dist2+N,INF); dist[0]=0; que.push(P(0,0)); while(!que.empty()){ P p=que.top(); que.pop(); int v=p.second,d=p.first; if(dist2[v]<d) continue; for(int i=0;i<G[v].size();i++){ edge e=G[v][i]; int d2=d+e.cost; if(dist[e.to]>d2){ swap(dist[e.to],d2); que.push(P(dist[e.to],e.to)); } if(dist2[e.to]>d2&&dist[e.to]<d2){ dist2[e.to]=d2; que.push(P(dist2[e.to],e.to)); } } } printf("%d\n",dist2[N-1]);}int main(){ scanf("%d%d",&N,&R); for(int i=0;i<R;i++){ int u,v,d; scanf("%d%d%d",&u,&v,&d); u--; v--; G[u].push_back((edge){v,d}); G[v].push_back((edge){u,d}); } solve(); return 0;}
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