poj2449 K短路模板题

昨晚看WC论文发现自己连K短路的经典A*算法还不会，补了一波，模板题输出-1后没return继续跑wa了一早上......

算法流程：

①在反向图中求出t到每个点的最短路

②从原点bfs，估价f=d+dis[x]，即当前已走的路径长度+最短路径

③遇到第k次汇点就是答案

据说这复杂度是O(n*k)的....不会证.....

代码：

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#define inf 1<<29
#define M 500200
#define N 500020
using namespace std;
inline int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int n,m,k,s,t;
int head
,pos;
int head1
,pos1;
struct edge{int to,c,next;}ee[M<<1],e[M<<1];
void add(int a,int b,int c)
{pos++;e[pos].to=b,e[pos].c=c,e[pos].next=head[a],head[a]=pos;}
void add1(int b,int a,int c)
{pos1++;ee[pos1].to=b,ee[pos1].c=c,ee[pos1].next=head1[a],head1[a]=pos1;}
queue<int>Q;bool vis
;int dis
;
void spfa(int st)
{
for(int i=1;i<=n;i++)
dis[i]=inf,vis[i]=0;
dis[st]=0,vis[st]=1;Q.push(st);
while(!Q.empty())
{
int u=Q.front();Q.pop();vis[u]=0;
for(int i=head1[u];i;i=ee[i].next)
{
int v=ee[i].to;
if(dis[v]>dis[u]+ee[i].c)
{
dis[v]=dis[u]+ee[i].c;
if(!vis[v])Q.push(v),vis[v]=1;
}
}
}
}
struct node{int x,d,f;}u,v;
bool operator < (node a,node b)
{
if(a.f!=b.f)return a.f>b.f;
}
priority_queue<node>que;
void solve()
{
while(!que.empty())que.pop();
if(dis[s]==inf)
{
printf("-1\n");
return ;
}
u.f=dis[s],u.x=s,u.d=0;
int now=0;que.push(u);
while(!que.empty())
{
u=que.top();que.pop();
if(u.x==t)
now++;
if(now==k)
{
printf("%d\n",u.d);
return;
}
for(int i=head[u.x];i;i=e[i].next)
{
int tc=e[i].to;v=u;
v.x=tc,v.d+=e[i].c;
v.f=dis[v.x]+v.d;
que.push(v);
}
}printf("-1\n");
}
int main()
{
// freopen("a.in","r",stdin);
// freopen("a.out","w",stdout);
while(~scanf("%d%d",&n,&m))
{
memset(head,0,sizeof(head));
memset(head1,0,sizeof(head1));
pos1=pos=0;
for(int i=1;i<=m;i++)
{
int x=read(),y=read();
int c=read();
add(x,y,c);
add1(x,y,c);
}s=read(),t=read(),k=read();
if(s==t)k++;
spfa(t);solve();
}
}