【模板】K短路 A-star

引理：当一个状态对应的节点第K次从堆中取出时，该状态对应的当前代价是从起点到该点的第K优解。

代码如下

/*
    POJ2449
*/
#include <cstdio>
#include <iostream>
#include <memory.h>
#include <queue>
#include <cstring>
#include <vector>
using namespace std;
const int maxv=1010;
const int maxe=1e5+10;

inline int read(){
    int x=0,f=1;char ch;
    do{ch=getchar();if(ch=='-')f=-1;}while(!isdigit(ch));
    do{x=x*10+ch-'0';ch=getchar();}while(isdigit(ch));
    return f*x;
}

struct edge{
    int nxt,to,w;
    edge(int x=0,int y=0,int z=0):nxt(x),to(y),w(z){}
}e[maxe],_e[maxe];
int tot=1,head[maxv],_tot=1,_head[maxv];

inline void add_edge(int from,int to,int w){
    e[++tot]=edge(head[from],to,w),head[from]=tot;
    _e[++_tot]=edge(_head[to],from,w),_head[to]=_tot;
}

int n,m,st,ed,k,d[maxv],cnt[maxv];
bool vis[maxv];

typedef pair<int,int> P;

void dij(){
    priority_queue<P> q;
    memset(d,0x3f,sizeof(d));
    d[ed]=0;q.push(make_pair(0,ed));
    while(q.size()){
        int u=q.top().second;q.pop();
        if(vis[u])continue;
        vis[u]=1;
        for(int i=_head[u];i;i=_e[i].nxt){
            int v=_e[i].to,w=_e[i].w;
            if(d[v]>d[u]+w){
                d[v]=d[u]+w;
                q.push(make_pair(-d[v],v));
            }
        }
    }
}

void read_and_parse(){
    n=read(),m=read();
    for(int i=1;i<=m;i++){
        int x=read(),y=read(),z=read();
        add_edge(x,y,z);
    }
    st=read(),ed=read(),k=read();
    if(st==ed)k++;
    dij();
}

struct node{//状态
    int idx,now,fur;
    node(int x=0,int y=0,int z=0):idx(x),now(y),fur(z){}
    friend bool operator<(const node& a,const node& b){
        return b.now+b.fur<a.now+a.fur;
    }
};

int solve(){
    priority_queue<node> q;
    q.push(node(st,0,d[st]));
    while(q.size()){
        node u=q.top();++cnt[u.idx];q.pop();
        if(cnt[u.idx]==k&&u.idx==ed)return u.now;
        if(cnt[u.idx]>k)continue;//如果当前节点出队次数大于K，则跳过
        for(int i=head[u.idx];i;i=e[i].nxt){
            int v=e[i].to,w=e[i].w;
            q.push(node(v,u.now+w,d[v]));
        }
    }
    return -1;
}

int main(){
    read_and_parse();
    printf("%d\n",solve());
    return 0;
}