题解 P4568 【[JLOI2011]飞行路线】
SuperJvRuo
2018-05-12 17:09:57
套路题,分层图。
以样例为例(使用 @EternalAlexander 这位dalao的OI Painter绘制):
![](https://cdn.luogu.com.cn/upload/pic/19106.png)
各层内部正常连边,各层之间从上到下连权值为0的边。每向下跑一层,就相当于免费搭一次飞机。跑一遍从$s$到$t+n*k$的最短路即可。
```
#include<cstdio>
#include<cctype>
#include<cstring>
#include<queue>
#include<algorithm>
#include<vector>
#include<utility>
#include<functional>
int Read()
{
int x=0;char c=getchar();
while(!isdigit(c))
{
c=getchar();
}
while(isdigit(c))
{
x=x*10+(c^48);
c=getchar();
}
return x;
}
using std::priority_queue;
using std::pair;
using std::vector;
using std::make_pair;
using std::greater;
struct Edge
{
int to,next,cost;
}edge[2500001];
int cnt,head[110005];
void add_edge(int u,int v,int c=0)
{
edge[++cnt]=(Edge){v,head[u],c};
head[u]=cnt;
}
int dis[110005];
bool vis[110005];
void Dijkstra(int s)
{
memset(dis,0x3f,sizeof(dis));
dis[s]=0;
priority_queue<pair<int,int>,vector<pair<int,int> >,greater<pair<int,int> > > points;
points.push(make_pair(0,s));
while(!points.empty())
{
int u=points.top().second;
points.pop();
if(!vis[u])
{
vis[u]=1;
for(int i=head[u];i;i=edge[i].next)
{
int to=edge[i].to;
if(dis[to]>dis[u]+edge[i].cost)
{
dis[to]=dis[u]+edge[i].cost;
points.push(make_pair(dis[to],to));
}
}
}
}
}
int main()
{
int n=Read(),m=Read(),k=Read(),s=Read(),t=Read();
int u,v,c;
for(int i=0;i<m;++i)
{
u=Read(),v=Read(),c=Read();
add_edge(u,v,c);
add_edge(v,u,c);
for(int j=1;j<=k;++j)
{
add_edge(u+(j-1)*n,v+j*n);
add_edge(v+(j-1)*n,u+j*n);
add_edge(u+j*n,v+j*n,c);
add_edge(v+j*n,u+j*n,c);
}
}
for(int i=1;i<=k;++i)
{
add_edge(t+(i-1)*n,t+i*n);
}//预防奇葩数据
Dijkstra(s);
printf("%d",dis[t+k*n]);
return 0;
}
```