poj 3255 Roadblocks
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此题为次短路问题,可以参考最短路问题的dijkstra算法,在这里使用了邻接表储存图,并使用优先队列优化,在dijkstra基础上开一个dis2数组储存次短路,分析次短路来源,有两种,一是dis[v]+cost[u][v],另外是dis2[v]+cost[u][v]。
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
/*13064844motefly3255Accepted4868K235MSG++1488B2014-07-12 16:13:01*/#include <iostream>#include <cstdio>#include <queue>#include <cstring>using namespace std;int n,r;typedef pair<int ,int > P;struct edge{ int from,to,cost;};const int INF=0x3f3f3f3f;const int MAXN=5000+10;vector<edge> G[MAXN];int dis[MAXN];int dis2[MAXN];void solve(){ priority_queue<P,vector<P>,greater<P> > que; memset(dis,INF,sizeof(dis)); memset(dis2,INF,sizeof(dis2)); dis[0]=0; que.push(P(0,0)); while(!que.empty()) { P p=que.top(); que.pop(); int v=p.second, d=p.first; if(dis2[v]<d) continue; for(int i=0;i<G[v].size();i++) { edge & e=G[v][i]; int d2=d+e.cost; if(dis[e.to]>d2) { swap(dis[e.to],d2); que.push(P(dis[e.to],e.to)); } if(dis2[e.to]>d2&&dis[e.to]<d2) { dis2[e.to]=d2; que.push(P(dis2[e.to],e.to)); } } } printf("%d\n",dis2[n-1]);}void init(){ scanf("%d%d",&n,&r); for(int i=0;i<r;i++) { int f,t,c; scanf("%d%d%d",&f,&t,&c); edge e; e.from=t-1; e.to=f-1; e.cost=c; G[t-1].push_back(e); e.from=f-1; e.to=t-1; e.cost=c; G[f-1].push_back(e); }}int main(){ init(); solve(); return 0;}
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