K短路-POJ2449

#include <queue>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <stdio.h>

using namespace std;
#define ll long long
#define mem(a, b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define DBG printf("this is a input\n")
#define fi first
#define se second
#define mk(a, b) make_pair(a,b)
#define p_queue priority_queue
struct e{
    int to , next , w;
}edge[200005],f_edge[200005];
int n , m, s, e, k;
p_queue <pair<int,int>,vector<pair<int,int> >,greater<pair<int,int> > >q;
int vis[1005] , dis[1005], cnt1 = 0, cnt2 = 0, head1[1005], head2[1005];
struct node
{
    int v ;
    int cost;
    bool operator <(const node&another)const{
        return cost + dis[v] > another.cost + dis[another.v];
    }
    node(int a,int c):v(a),cost(c) {}
};
p_queue <node> aq;
void add1(int f, int t, int w)
{
    edge[cnt1].to = t;
    edge[cnt1].w = w;
    edge[cnt1].next = head1[f];
    head1[f] = cnt1 ++;
}
void add2(int f, int t, int w)
{
    f_edge[cnt2].to = t;
    f_edge[cnt2].w = w;
    f_edge[cnt2].next = head2[f];
    head2[f] = cnt2 ++;
}
void dijsktra()
{
    pair<int,int> pa;
    pa.first = 0;
    pa.second = e;
    dis[e] = 0;
    q.push(pa);
    while(!q.empty())
    {
        pair<int,int> no = q.top();
        q.pop();
        if(vis[no.second]) continue;
        vis[no.second] = 1;
        for(int i = head2[no.second] ; i != -1 ; i = f_edge[i].next)
        {
            int v = f_edge[i].to ,w = f_edge[i].w;
            if(dis[no.second] + w < dis[v] && !vis[v])
            {
                dis[v] = dis[no.second] + w;
                q.push(mk(dis[v],v));
            }
        }
    }
}
int Astar()
{
    aq.push(node(s,0));
    while(!aq.empty())
    {
        node no = aq.top();
        aq.pop();
        if(no.v == e)
        {
            k--;
            if(k == 0)
                return no.cost;
        }
        for(int i = head1[no.v] ; i != -1 ; i = edge[i].next)
            aq.push(node(edge[i].to,no.cost+edge[i].w));
    }
    return -1;
}
int main(void) {
    mem(dis,INF);
    mem(head1,-1);
    mem(head2,-1);
    mem(vis,0);
    scanf("%d %d",&n, &m);
    for(int i = 1 ; i <= m ; i ++)
    {
        int u ,v , w;
        scanf("%d %d %d",&u, &v, &w);
        add1(u,v,w);
        add2(v,u,w);
    }
    scanf("%d %d %d",&s, &e, &k);
    if(s == e)
        k ++;
    dijsktra(); //先求出所有点到终点的距离（反向建图，终点作为起点跑dij即可）
    printf("%d\n",Astar());//A*算法用于求出k短路
}

k短路求解步骤：

1.建立反向图，并求出任意点到终点的最短距离（dij算法）

2.利用A*进行搜索 启发式函数：

对于A*,估价函数=当前值+当前位置到终点的距离,即f(V)=g(V)+h(V),每次扩展估价函数值最小的一个;

对于K短路算法来说:

                                g(V)为当前从S到V所走的路径的长度（广搜变量记录）;

                                h(V)为点V到的最短路的长度（第一步已经求出）;