poj-3255-Roadblocks-路径可重复次短路

题目：

Roadblocks
Time Limit: 2000MS        Memory Limit: 65536K
Total Submissions: 7075        Accepted: 2629
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)

题目

题意：无向图。 起点为1， 终点为 n, 输出仅次于 最短路的次短路， 路径可重复走。

思路：正反求两遍最短路，得出1 到所有点的最短距离， 和 n到所有点的最短距离。因为路径可以重复走，所以两点之间只有一条边的可以重复经过。接着将每条边视作单项边，枚举每条单项边e(u, v)，Dist = dist(1, u) + dist(n, v) + dist(u,v).这样只会重复一条最小边。所求得的 大于 dist(1, n)的最小的Dist，就是题目的解。（理解不了的画一下图就明白了）

AC代码：

 #include<cstdio>
 #include<iostream>
 #include<queue>
 #include<vector>
 #include<algorithm>
 #include<cstring>
 #include<climits>
 using namespace std;
 #define maxn 100009
 #define INF INT_MAX-100000
 struct Edge
 {
    int from, to, dist;
 };

 struct Heapnode
 {
    int d, u;
    bool operator < (const Heapnode& that) const {
       return d > that.d;
    }
 };

 int dist1[maxn], dist2[maxn];
 struct Dijkstra
 {
    int n, m;
    vector<Edge> edges;
    vector<int> G[maxn];
    bool done[maxn];
    int d[maxn];
    int p[maxn];

    void init(int n) {
       this->n = n;
       for(int i = ; i < n; i++) G[i].clear();
       edges.clear();
    }

    void AddEdge(int from, int to , int dist) {
       edges.push_back((Edge) {from, to , dist});
       m = edges.size();
       G[from].push_back(m-);
    }

    void dijkstra(int s, int* dx) {
       priority_queue<Heapnode> Q;
       for(int i = ; i < n; i++) dx[i] = d[i] = INF;
       d[s] = dx[s] = ;
       memset(done, , sizeof(done));
       Q.push((Heapnode) {, s});
       while(!Q.empty()) {
          Heapnode x = Q.top(); Q.pop();
          int u = x.u;
          if(done[u]) continue;
          done[u] = true;
          int size = G[u].size();
          for(int i = ; i < size; i++) {
             Edge& e = edges[G[u][i]];
             if(d[e.to] > d[u] + e.dist) {
                dx[e.to] = d[e.to] = d[u] + e.dist;
                p[e.to] = G[u][i];
                Q.push ((Heapnode) {d[e.to], e.to});
             }
          }
       }
    }
 }dij;

 void work(int n, int r)
 {
    dij.init(n+);
    for(int i = ; i < r; i++){
       int a, b, c; scanf("%d%d%d", &a, &b, &c);
       dij.AddEdge(a, b, c);
       dij.AddEdge(b, a, c);
    }
    dij.dijkstra(, dist1);
    dij.dijkstra(n, dist2);
    int Dist, sta = dist1[n], res = INF;
    for(int i = ; i <= n; i++){
       int ss = dij.G[i].size();
       for(int j = ; j < ss; j++){
          int m = dij.G[i][j];
          Edge ee = dij.edges[m];
          Dist = dist1[ee.from] + dist2[ee.to] + ee.dist;
          if(Dist < res && Dist > dist1[n]) res = Dist;
          //cout<<Dist<<" "<<res<<endl;
       }
    }
    printf("%d\n", res);
 }
 int main()
 {
    int n, r;
    while(scanf("%d%d", &n, &r) != EOF){
       work(n, r);
    }
    return ;
 }