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).

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)

Line 1: The length of the second shortest path between node 1 and node N

4 4
1 2 100
2 4 200
2 3 250
3 4 100

450

Two routes: 1 -> 2 -> 4 (length 100+200=300) and 1 -> 2 -> 3 -> 4 (length 100+250+100=450)

 

 

题意就是要求次短路的长度，这道题是根据POJ2449而来的，因为2449求的是k短路，如果任意的第K短路都能求了，那么次短路还在话下吗？

 

#include <stdio.h>
#include <string.h>
#include <queue>
#include <algorithm>
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.g<g;
        return a.f<f;
    }
};
int head[l],vis[l],dis[l];

void init()
{
    memset(e,-1,sizeof(e));
    memset(head,-1,sizeof(head));
    memset(vis,0,sizeof(vis));
    for(int i = 0; i<=n; i++)
        dis[i] = inf;
}

void AddEdges(int x,int y,int w,int k)
{
    e[k].x = x,e[k].y = y,e[k].w = w,e[k].next = head[x],head[x] = k;
}

int relax(int u,int v,int c)
{
    if(dis[v]>dis[u]+c)
    {
        dis[v] = dis[u]+c;
        return 1;
    }
    return 0;
}

void spfa(int src)
{
    int a,next,i;
    queue<int> 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<node> 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;
}