题目链接

题意：给你一个图以及起点和终点，求起点到终点的第k短路的大小。

思路：先用迪杰斯特拉求出每个点到终点的最短路，然后用A*算法，令f(x)为每点到终点的距离，将f(x)附加到每条边的边权上，再利用优先队列对最小的进行扩展，直接暴力搜相邻边，则第k个从优先队列中出来的就是正解。思路很明了，代码写起来很复杂。。。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#define ll long long
using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = 1005;
struct edge
{
    int v, w, nxt;
}G[100005], G2[100005];
int tot, pre[maxn], t2, p2[maxn];
struct node 
{
    int v, d;
    friend bool operator < (node a, node b)     
    {         
        if(a.d!=b.d)
            return a.d>b.d;
        return a.v>b.v; 
    }
};
int n, m, k, dist[maxn];
bool vis[maxn];
void dijkstra(int s) 
{
    memset(dist, 0x3f, sizeof(dist));
    memset(vis, 0, sizeof(vis));
    dist[s] = 0;
    priority_queue<node> que;
    node y;
    y.v=s;
    y.d=0;
    que.push(y);
    while (!que.empty())
    {
        node p = que.top();
        que.pop();
        if (vis[p.v])
        {
            continue;
        }
        vis[p.v] = true;
        for (int i = p2[p.v]; ~i; i = G2[i].nxt)
        {
            int v = G2[i].v, w = G2[i].w;
            if (!vis[v] && p.d + w < dist[v])
            {
                dist[v] = p.d + w;
                node x;
                x.v=v;
                x.d=dist[v];
                que.push(x);
            }
        }
    }
}
struct point 
{
    int v, h, g;
    friend bool operator < (point a, point b)     
    {         
        return a.h+a.g>b.h+b.g; 
    }
};
int times[maxn];
int Astar(int s, int e) 
{
    if(dist[s]==inf)
    {
        return -1;
    }
    memset(times, 0, sizeof(times));
    priority_queue<point> Q;
    point y;
    y.v=s;
    y.g=0;
    y.h=0;
    Q.push(y);
    while (!Q.empty()) {
        point p = Q.top();
        Q.pop();
        ++times[p.v];
        if (times[p.v] == k && p.v == e)
        {
            return p.h + p.g;
        }
        if (times[p.v] > k) {
            continue;
        }
        for (int i = pre[p.v]; ~i; i = G[i].nxt)
        {
            point x;
            x.v=G[i].v;
            x.h=p.h+G[i].w;
            x.g=dist[G[i].v];
            Q.push(x);
        }
    }
    return -1;
}
int main() {
    int u, v, w, s, e;
    scanf("%d%d", &n, &m);
    tot = t2 = 0;
    memset(pre, -1, sizeof(pre));
    memset(p2, -1, sizeof(p2));
    while(m--)
    {
        scanf("%d%d%d", &u, &v, &w);
        G[tot].v = v;
        G[tot].w = w;
        G[tot].nxt = pre[u];
        pre[u] = tot++;
        G2[t2].v = u;
        G2[t2].w = w;
        G2[t2].nxt = p2[v];
        p2[v] = t2++;
    }
    scanf("%d%d%d", &s, &e, &k);
    if (s == e)
    {
        k++;
    }
    dijkstra(e);
    printf("%d\n", Astar(s, e));
    return 0;
}