POJ3255:Roadblocks(次短路 SPFA+A星)

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) 題意就是要求次短路的長度，這道題是根據POJ2449而來的，因為2449求的是k短路，如果任意的第K短路都能求了，那麼次短路還在話下嗎？

#include 
#include 
#include 
#include 
using namespace std;
const int L = 100005;
const int l = 5005;
const int inf = 1<<30;

int n,m;
struct edges
{
    int x,y,w,next;
} e[L<<2];

struct node
{
    int now,g,h,f;
    bool operator<(const node a)const
    {
        if(a.f == f) return a.gdis[u]+c)
    {
        dis[v] = dis[u]+c;
        return 1;
    }
    return 0;
}

void spfa(int src)
{
    int a,next,i;
    queue Q;
    Q.push(src);
    dis[src] = 0;
    while(!Q.empty())
    {
        a = Q.front();
        Q.pop();
        vis[a] = 0;
        for(i = head[a]; i!=-1; i = e[i].next)
        {
            next = e[i].y;
            if(relax(a,next,e[i].w) && !vis[next])
            {
                Q.push(next);
                vis[next] = 1;
            }
        }
    }
}

void Astar(int src,int to)
{
    int i,cnt = 0;
    priority_queue Q;
    node a,next;
    a.now = src;
    a.g = 0;
    a.f = a.g+dis[src];
    Q.push(a);
    while(!Q.empty())
    {
        a = Q.top();
        Q.pop();
        if(a.now == to)
        {
            cnt++;
            if(cnt == 2)
            {
                printf("%d\n",a.g);
                return ;
            }
        }
        for(i = head[a.now]; i!=-1; i = e[i].next)
        {
            next = a;
            next.now = e[i].y;
            next.g = a.g+e[i].w;
            next.f = next.g+dis[next.now];
            Q.push(next);
        }
    }
}

int main()
{
    int i,j,x,y,w;
    while(~scanf("%d%d",&n,&m))
    {
        init();
        for(i = 0; i<2*m; i+=2)
        {
            scanf("%d%d%d",&x,&y,&w);
            AddEdges(x,y,w,i);
            AddEdges(y,x,w,i+1);
        }
        spfa(n);
       Astar(1,n);
    }

    return 0;
}