标签：code start ati .net pst test tput 设计 tar
| Time Limit: 2000MS | Memory Limit: 65536K | |
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
Output
Sample Input
4 4
1 2 100
2 4 200
2 3 250
3 4 100
Sample Output
450
Hint
#include <iostream> #include <cstdio> #include <vector> #include <queue> using namespace std; const int MAX_V=5001; const int INF=1e9; struct edge{ int to;//边的出点 int cost;//权值 }; vector<edge> G[MAX_V];//邻接表 typedef pair<int,int> P;//first是源点到该点的距离，second是当前顶点 int dist1[MAX_V];//存最短距离 int dist2[MAX_V];//存次短距离 int N,R;//N个路口,R条道路 void dijkstra(); int main() { int s,w,t; edge e; scanf("%d%d",&N,&R); for(int i=0;i<R;i++){ scanf("%d%d%d",&s,&t,&w); e.to=t-1; e.cost=w; G[s-1].push_back(e); e.to=s-1; G[t-1].push_back(e); } dijkstra(); printf("%d\n",dist2[N-1]); return 0; } void dijkstra(){ priority_queue <P,vector<P>,greater<P> >que; fill(dist1,dist1+N,INF); fill(dist2,dist2+N,INF); dist1[0]=0; que.push(P(0,0)); while(!que.empty()){ P p=que.top(); que.pop(); int v=p.second;//新的源点 int d=p.first; /*如果旧源点到新源点的距离比旧源点到新源点的距离大， 那么不用执行下面的代码去更新新源点到其他各点的dist1,dist2 因为它算出来的距离肯定比以前算出来的大。 */ if(d>dist2[v]) continue; for(int i=0;i<G[v].size();i++){ edge& e=G[v][i]; int d2=d+e.cost; if(dist1[e.to]>d2){///最短距离 swap(d2,dist1[e.to]); que.push(P(dist1[e.to], e.to)); } /* d2大于源点到e.to的最短距离，小于以前计算的次短距离则更新 */ if(d2>dist1[e.to]&&d2<dist2[e.to]){///次短距离 dist2[e.to]=d2; que.push(P(dist2[e.to],e.to)); } } } }
参考自http://blog.csdn.net/gemire/article/details/20832199
标签：code start ati .net pst test tput 设计 tar
原文地址：http://www.cnblogs.com/LuRenJiang/p/7419588.html
