BZOJ 1598 第k短路

思路：
先反向建图 Dijkstra一遍 求出h数组
再正向建图 A_star一遍 搞定

//By SiriusRen#include <queue>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define int long long#define N 10050int first[N],next[N],v[N],w[N],tot,h[N],vis[N],ans[N];int n,m,k,xx[N],yy[N],zz[N];struct Node{int h,g,now;}jy;priority_queue<Node>pq;void add(int x,int y,int z){w[tot]=z,v[tot]=y,next[tot]=first[x],first[x]=tot++;}bool operator < (Node a,Node b){    return a.h+a.g>b.h+b.g;}void Dijkstra(){    jy.now=1,pq.push(jy);    for(int i=2;i<=n;i++)h[i]=0x3ffffff;    while(!pq.empty()){        int x=pq.top().now;pq.pop();        if(vis[x])continue;        vis[x]=1;        for(int i=first[x];~i;i=next[i])            if(h[v[i]]>h[x]+w[i]){                h[v[i]]=h[x]+w[i];                jy.h=h[v[i]],jy.now=v[i];                pq.push(jy);            }    }}void A_star(){    jy.h=jy.g=0,jy.now=n;pq.push(jy);    while(!pq.empty()){        Node t=pq.top();pq.pop();        if(vis[t.now]>k)continue;        if(t.now==1)ans[vis[1]]=t.g;        vis[t.now]++;        for(int i=first[t.now];~i;i=next[i]){            jy.h=h[v[i]],jy.g=t.g+w[i],jy.now=v[i];            pq.push(jy);        }    }}signed main(){    memset(first,-1,sizeof(first));    scanf("%lld%lld%lld",&n,&m,&k);    for(int i=1;i<=m;i++){        scanf("%lld%lld%lld",&xx[i],&yy[i],&zz[i]);        add(yy[i],xx[i],zz[i]);    }    Dijkstra();    memset(first,-1,sizeof(first)),tot=0;    for(int i=1;i<=m;i++)add(xx[i],yy[i],zz[i]);    A_star();    for(int i=1;i<vis[1];i++)printf("%lld\n",ans[i]);    for(int i=vis[1];i<=k;i++)puts("-1");}