Roadblocks Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 10618 Accepted: 3772 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 4 1 2 100 2 4 200 2 3 250 3 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 错因：没有熟练掌握STL的应用 解答：1：核心算法：到某点v的次短路只能是到另外一点u的最短路再加上u到v的权值cost[u][v] 因此需要设置两个数组dist[maxn]和dist2[maxn]分别代表达到该点的最短路和次短路 2：易错点：如代码所示不能将dist2[0]的初值也一同设为0，比如假设只有两个点，权值为1，那么次短路显然只能为2，而不是0.。。 3：需要掌握的知识点： STL知识： （1）.pair的优先队列的使用以及排序 typedef pair P; priority_queue ,greater
> q; （2）.赋初值fill函数的使用 fill(dist,dist+5005,inf); （3）.邻接表的使用 vector
G[5005]; #include #include #include #include #include #include #include using namespace std; #define inf 1e9 typedef pair P; struct edge{ int t,c; }; vector G[5005]; int dist[5005],dist2[5005]; int main() { int n,r; while(~scanf("%d %d",&n,&r)) { fill(dist,dist+5005,inf); fill(dist2,dist2+5005,inf); priority_queue ,greater
> q; int x; edge w; dist[0]=0;//2 赋初值dist2[0]=0将出错。 for(int i=1;i<=r;i++) {
scanf("%d %d %d",&x,&w.t,&w.c); x--; w.t--; G[x].push_back(w); swap(w.t,x); G[x].push_back(w); } q.push(P(0,0)); while(!q.empty()) { P p=q.top(); q.pop(); int v=p.second,d=p.first; if(d>dist2[v]) continue; for(int i=0; i d2) { swap(dist[e.t],d2); q.push(P(dist[e.t],e.t)); } if(dist2[e.t]>d2&&dist[e.t]