background
\ (USACO \) \ (2009 \) \ (On Feb \) \ (T3 \) , \ (Luogu \) \ (P2939 / BZOJ1579 \)
The meaning of problems
Given \ (n-\) points, \ (m \) two end edges of the \ (X, Y \) , the path may be such that a predetermined maximum of \ (K \) edges becomes \ (0 \) . Seeking \ (1 \) to the \ (n-\) of the shortest path length.
solution
Figure layered shortest template.
Since the free \ (k \) times, built \ (k \) layer diagram (assuming that becomes \ (0 \) , the more the higher the number of edges in the graph) can be.
Note that each edge connected again at each level, and \ (X \) to a higher-level \ (Y \) connected \ (0 \) side, \ (Y \) to a higher-level \ (X \ ) even \ (0 \) side.
From \ (1 \) No. o'clock \ (Dijkstra \) , the answer is first of all layers \ (n-\) Shortest Path minimum points.
detail
Please be sure to pay attention to check the map topic array bloom big enough . Make sure sure sure substituted into the limit data validation .
Code
\(View\) \(Code\)
#include<bits/stdc++.h>
using namespace std;
inline int read()
{
int ret=0,f=1;
char ch=getchar();
while(ch>'9'||ch<'0')
{
if(ch=='-')
f=-1;
ch=getchar();
}
while(ch>='0'&&ch<='9')
{
ret=(ret<<1)+(ret<<3)+ch-'0';
ch=getchar();
}
return ret*f;
}
const int inf=0x3f3f3f3f;
int n,m,k,s,t,u,v,w;
int dis[220005],ans=inf;
int num,head[220005];
bool vis[220005];
struct edge
{
int ver,nxt,w;
}e[4100005];
inline void adde(int u,int v,int w)
{
e[++num].ver=v;
e[num].w=w;
e[num].nxt=head[u];
head[u]=num;
}
void dijkstra(int s)
{
memset(dis,0x3f,sizeof(dis));
memset(vis,0,sizeof(vis));
priority_queue<pair<int,int> > q;
dis[s]=0;
q.push(make_pair(0,s));
while(!q.empty())
{
int x=q.top().second;
q.pop();
if(vis[x])
continue;
vis[x]=1;
for(register int i=head[x];i;i=e[i].nxt)
{
int y=e[i].ver,w=e[i].w;
if(dis[x]+w<dis[y])
{
dis[y]=dis[x]+w;
q.push(make_pair(-dis[y],y));
}
}
}
}
int main()
{
n=read();
m=read();
k=read();
s=1;
t=n;
for(register int i=1;i<=m;i++)
{
u=read();
v=read();
w=read();
adde(u,v,w);
adde(v,u,w);
for(register int j=1;j<=k;j++)
{
adde(u+(j-1)*n,v+j*n,0);
adde(v+(j-1)*n,u+j*n,0);
adde(u+j*n,v+j*n,w);
adde(v+j*n,u+j*n,w);
}
}
dijkstra(s);
for(register int i=0;i<=k;i++)
ans=min(ans,dis[t+i*n]);
if(ans<inf)
printf("%d\n",ans);
else
printf("-1\n");
return 0;
}