POJ2449 Remmarguts' Date 第K短路

POJ2449 比较裸的K短路问题

K短路听起来高大上 实际思路并不复杂

首先对终点t到其他所有点求最短路 即为dist[]

然后由起点s 根据当前走过的距离+dist[]进行A*搜索 第k次到达t即为第K短路

代码也很简单

//数组开的不够 不一定是运行时错误！ 可能也会WA 
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<vector>
using namespace std;
typedef long long int LL;
const int maxn=(1e+3)+1,maxm=(1e+5)+1;
vector<int > son[maxn];vector<int >fa[maxn];
int ea[maxm],eb[maxm],len[maxm];
LL dist[maxn];
bool vis[maxn];
typedef pair<int,int>po;
const LL INF=1e+18;
priority_queue<po,vector<po>,greater<po> >q;
int times;
po p[maxn];
struct pi
{
 LL f;
 LL dis;
 int th;
 pi(LL a,LL b,int c)
 	{
 	 f=a;
 	 dis=b;
 	 th=c;
	}
 bool operator<(const pi b)const
 	{
	 if(f==b.f)return dis>b.dis; 
 	 return f>b.f;
	}	
};
priority_queue<pi>Q;
int main()
{freopen("t.txt","r",stdin);
 int n,m;
 scanf("%d%d",&n,&m);
 for(int i=1;i<=n;i++)
 	{son[i].clear();fa[i].clear();}
 for(int i=1;i<=m;i++)
 	{
 	 int a,b,l;	
     scanf("%d%d%d",&a,&b,&l);
	 ea[i]=a;eb[i]=b;len[i]=l;
	 son[a].push_back(i);fa[b].push_back(i);		
	}	
 int s,t,k;
 scanf("%d%d%d",&s,&t,&k);
 for(int i=1;i<=n;i++)
  	{dist[i]=INF;}
 while(!q.empty())q.pop();
 dist[t]=0;
 for(int i=1;i<=n;i++)
 	p[i]=po(dist[i],i);
 q.push(p[t]);
 memset(vis,0,sizeof(vis));
 while(!q.empty())
 	{
 	 int mark;
	 po now=q.top();
	 q.pop();	
 	 mark=now.second;
	 if(vis[mark])continue;
	 vis[mark]=true;
	 for(int i=0;i<fa[mark].size();i++)
	 	{
	 	 int e=fa[mark][i];
	 	 int father=ea[e];	
	 	 if(dist[father]>(dist[mark]+len[e]))
		    {dist[father]=dist[mark]+len[e];po np(dist[father],father);q.push(np);}
				
		}	
	}
 while(!Q.empty())Q.pop();
 pi start(dist[s],0,s);
 Q.push(start);
 LL ans=-1;
 if(s==t)k++;
 while(!Q.empty())
 	{
 	 if(dist[s]==INF)break;
 	 pi now=Q.top();	
 	 Q.pop();
	 int npo=now.th;
	 if(npo==t)	times++;
	 if(npo==t&&times==k){ans=now.dis;break;}
	 for(int i=0;i<son[npo].size();i++)
	 	{
	 	 int e=son[npo][i];
	 	 int so=eb[e];
	 	 pi ps(now.dis+len[e]+dist[so],now.dis+len[e],so);
	 	 Q.push(ps);
		}
	}
 cout<<ans<<endl;
 return 0;	
}