poj2449

题意：给定一个图，求起点到终点的第k短路。

分析：先用dijkstra从t反向寻找最短路。然后使用A*算法，把f(i)=g(i) + h(i)。h(i)就是i点到t的最短距离。当某点出队次数达到k次的时候，结果为该点的当前路程＋该点到t的最短距离。（我没有判断不连通的情况）

为什么这样做是对的呢？我们这样来思考，如果不实用最短路，而只使用A*那么t第x次出队的结果即为第x短路的距离。继而可以想到，从第一个出队次数达到x的点，沿着最短路走到t，一定是第x短路。

说实话我也没有完全理解。

另外注意s==t的情况，据说k要＋＋，不明白为啥。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <queue>
usingnamespace std;

#define V 1005
#define E 1000005

int n, m, s, t, k;
#define typec int
// type of cost
const typec inf =0x3f3f3f3f;
// max of cost
typec cost[E], dist[V];
int e, pnt[E], nxt[E], head[V], prev[V], vis[V];
int pnt2[E], nxt2[E], head2[V], prev2[V], cost2[E];

struct pqnode
{
    int v, dist;
    pqnode(int vv =0, typec cc =0) :
        v(vv), dist(cc)
    {
    }
};

struct qnode
{
    int v;
    typec c;
    qnode(int vv =0, typec cc =0) :
        v(vv), c(cc)
    {
    }
    booloperator<(const qnode& r) const
    {
        return c > r.c;
    }
};

booloperator<(const pqnode &a, const pqnode &b)
{
    return a.dist + dist[a.v] > b.dist + dist[b.v];
}

void dijkstra(int n, constint src)
{
    qnode mv;
    int i, j, k, pre;
    priority_queue<qnode> que;
    vis[src] =1;
    dist[src] =0;
    que.push(qnode(src, 0));
    for (pre = src, i =1; i < n; i++)
    {
        for (j = head[pre]; j !=-1; j = nxt[j])
        {
            k = pnt[j];
            if (vis[k] ==0&& dist[pre] + cost[j] < dist[k])
            {
                dist[k] = dist[pre] + cost[j];
                que.push(qnode(pnt[j], dist[k]));
                prev[k] = pre;
            }
        }
        while (!que.empty() && vis[que.top().v] ==1)
            que.pop();
        if (que.empty())
            break;
        mv = que.top();
        que.pop();
        vis[pre = mv.v] =1;
    }
}

inline void addedge(int u, int v, typec c)
{
    pnt[e] = v;
    cost[e] = c;
    nxt[e] = head[u];
    head[u] = e++;
}

inline void addedge2(int u, int v, typec c)
{
    pnt2[e] = v;
    cost2[e] = c;
    nxt2[e] = head2[u];
    head2[u] = e++;
}

void input()
{
    memset(head, -1, sizeof(head));
    memset(vis, 0, sizeof(vis));
    memset(prev, -1, sizeof(prev));
    memset(head2, -1, sizeof(head2));
    memset(prev2, -1, sizeof(prev2));
    scanf("%d%d", &n, &m);
    for (int i =0; i < n; i++)
        dist[i] = inf;
    for (int i =0; i < m; i++)
    {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        a--;
        b--;
        addedge2(a, b, c);
        addedge(b, a, c);
    }
    scanf("%d%d%d", &s, &t, &k);
    s--;
    t--;
    if (s == t)
        k++;
}

int astar()
{
    memset(vis, 0, sizeof(vis));
    priority_queue<pqnode> pq;
    pq.push(pqnode(s, 0));
    while (!pq.empty())
    {
        pqnode pre = pq.top();
        pq.pop();
        vis[pre.v]++;
        if (vis[pre.v] == k)
            return pre.dist + dist[pre.v];
        for (int i = head2[pre.v]; i !=-1; i = nxt2[i])
            if (vis[pnt2[i]] <= k)
                pq.push(pqnode(pnt2[i], pre.dist + cost2[i]));
    }
    return-1;
}

int main()
{
//    freopen("t.txt", "r", stdin);
    input();
    dijkstra(n, t);
    printf("%d\n", astar());
    return0;
}