RoadblocksTime Limit: 2000MS        Memory Limit: 65536KTotal Submissions: 7075        Accepted: 2629DescriptionBessie 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).InputLine 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)OutputLine 1: The length of the second shortest path between node 1 and node NSample Input4 41 2 1002 4 2002 3 2503 4 100Sample Output450HintTwo routes: 1 -> 2 -> 4 (length 100+200=300) and 1 -> 2 -> 3 -> 4 (length 100+250+100=450)

  1 #include<cstdio>  2 #include<iostream>  3 #include<queue>  4 #include<vector>  5 #include<algorithm>  6 #include<cstring>  7 #include<climits>  8 using namespace std;  9 #define maxn 100009 10 #define INF INT_MAX-100000 11 struct Edge 12 { 13    int from, to, dist; 14 }; 15  16 struct Heapnode 17 { 18    int d, u; 19    bool operator < (const Heapnode& that) const { 20       return d > that.d; 21    } 22 }; 23  24 int dist1[maxn], dist2[maxn]; 25 struct Dijkstra 26 { 27    int n, m; 28    vector<Edge> edges; 29    vector<int> G[maxn]; 30    bool done[maxn]; 31    int d[maxn]; 32    int p[maxn]; 33  34    void init(int n) { 35       this->n = n; 36       for(int i = 0; i < n; i++) G[i].clear(); 37       edges.clear(); 38    } 39  40    void AddEdge(int from, int to , int dist) { 41       edges.push_back((Edge) {from, to , dist}); 42       m = edges.size(); 43       G[from].push_back(m-1); 44    } 45  46    void dijkstra(int s, int* dx) { 47       priority_queue<Heapnode> Q; 48       for(int i = 0; i < n; i++) dx[i] = d[i] = INF; 49       d[s] = dx[s] = 0; 50       memset(done, 0, sizeof(done)); 51       Q.push((Heapnode) {0, s}); 52       while(!Q.empty()) { 53          Heapnode x = Q.top(); Q.pop(); 54          int u = x.u; 55          if(done[u]) continue; 56          done[u] = true; 57          int size = G[u].size(); 58          for(int i = 0; i < size; i++) { 59             Edge& e = edges[G[u][i]]; 60             if(d[e.to] > d[u] + e.dist) { 61                dx[e.to] = d[e.to] = d[u] + e.dist; 62                p[e.to] = G[u][i]; 63                Q.push ((Heapnode) {d[e.to], e.to}); 64             } 65          } 66       } 67    } 68 }dij; 69  70 void work(int n, int r) 71 { 72    dij.init(n+1); 73    for(int i = 0; i < r; i++){ 74       int a, b, c; scanf("%d%d%d", &a, &b, &c); 75       dij.AddEdge(a, b, c); 76       dij.AddEdge(b, a, c); 77    } 78    dij.dijkstra(1, dist1); 79    dij.dijkstra(n, dist2); 80    int Dist, sta = dist1[n], res = INF; 81    for(int i = 1; i <= n; i++){ 82       int ss = dij.G[i].size(); 83       for(int j = 0; j < ss; j++){ 84          int m = dij.G[i][j]; 85          Edge ee = dij.edges[m]; 86          Dist = dist1[ee.from] + dist2[ee.to] + ee.dist; 87          if(Dist < res && Dist > dist1[n]) res = Dist; 88          //cout<<Dist<<" "<<res<<endl; 89       } 90    } 91    printf("%d\n", res); 92 } 93 int main() 94 { 95    int n, r; 96    while(scanf("%d%d", &n, &r) != EOF){ 97       work(n, r); 98    } 99    return 0;100 }