描述
http://www.lydsy.com/JudgeOnline/problem.php?id=1614
分析
类似POJ_3662_Telephone_Lines_(二分+最短路)
Dijkstra:
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 const int maxn=1000+5,maxm=10000+5,INF=1<<30; 5 int n,m,k,ect; 6 int hd[maxn],f[maxm],d[maxn]; 7 bool vis[maxn]; 8 struct edge{ 9 int to,w,next; 10 edge(int to=0,int w=0,int next=0):to(to),w(w),next(next){} 11 bool operator < (const edge &a) const { return w>a.w; } 12 }g[maxm<<1]; 13 inline int read(int &x){ x=0;int k=1;char c;for(c=getchar();c<'0'||c>'9';c=getchar())if(c=='-')k=-1;for(;c>='0'&&c<='9';c=getchar())x=x*10+c-'0';return x*=k; } 14 inline void add_edge(int u,int v,int w){ 15 g[++ect]=edge(v,w,hd[u]); hd[u]=ect; 16 g[++ect]=edge(u,w,hd[v]); hd[v]=ect; 17 } 18 inline bool C(int x){ 19 for(int i=1;i<=n;i++) d[i]=INF, vis[i]=false; 20 d[1]=0; 21 priority_queue <edge> q; 22 q.push(edge(1,0,0)); 23 while(!q.empty()){ 24 int u=q.top().to; q.pop(); 25 if(vis[u]) continue; 26 vis[u]=true; 27 for(int i=hd[u];i;i=g[i].next){ 28 int v=g[i].to,duv=g[i].w>x?1:0; 29 if(d[v]>d[u]+duv){ 30 d[v]=d[u]+duv; 31 q.push(edge(v,d[v],0)); 32 } 33 } 34 } 35 return d[n]<=k; 36 } 37 inline int bsearch(int l,int r){ 38 if(!C(f[r])) return-1; 39 while(l<r){ 40 int mid=l+(r-l)/2; 41 if(C(f[mid])) r=mid; 42 else l=mid+1; 43 } 44 return f[l]; 45 } 46 int main(){ 47 read(n); read(m); read(k); 48 for(int i=1,u,v,w;i<=m;i++){ 49 read(u); read(v); read(w); 50 add_edge(u,v,w); 51 f[i]=w; 52 } 53 sort(f+1,f+1+m); 54 printf("%d ",bsearch(0,m)); 55 return 0; 56 }
Spfa:
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 const int maxn=1000+5,maxm=10000+5,INF=1<<30; 5 int n,m,k,ect; 6 int hd[maxn],q[maxn],f[maxm],d[maxn]; 7 bool vis[maxn]; 8 struct edge{ 9 int to,w,next; 10 edge(int to=0,int w=0,int next=0):to(to),w(w),next(next){} 11 }g[maxm<<1]; 12 inline int read(int &x){ x=0;int k=1;char c;for(c=getchar();c<'0'||c>'9';c=getchar())if(c=='-')k=-1;for(;c>='0'&&c<='9';c=getchar())x=x*10+c-'0';return x*=k; } 13 inline void add_edge(int u,int v,int w){ 14 g[++ect]=edge(v,w,hd[u]); hd[u]=ect; 15 g[++ect]=edge(u,w,hd[v]); hd[v]=ect; 16 } 17 inline bool C(int x){ 18 for(int i=1;i<=n;i++) d[i]=INF, vis[i]=false; 19 d[1]=0,vis[1]=true; 20 int l=0,r=0; 21 q[r++]=1; 22 while(l!=r){ 23 int u=q[l++]; if(l==maxn-1) l=0; 24 vis[u]=false; 25 for(int i=hd[u];i;i=g[i].next){ 26 int v=g[i].to,duv=g[i].w>x?1:0; 27 if(d[v]>d[u]+duv){ 28 d[v]=d[u]+duv; 29 if(!vis[v]){ 30 q[r++]=v; if(r==maxn-1) r=0; 31 vis[v]=true; 32 } 33 } 34 } 35 } 36 return d[n]<=k; 37 } 38 inline int bsearch(int l,int r){ 39 if(!C(f[r])) return-1; 40 while(l<r){ 41 int mid=l+(r-l)/2; 42 if(C(f[mid])) r=mid; 43 else l=mid+1; 44 } 45 return f[l]; 46 } 47 int main(){ 48 read(n); read(m); read(k); 49 for(int i=1,u,v,w;i<=m;i++){ 50 read(u); read(v); read(w); 51 add_edge(u,v,w); 52 f[i]=w; 53 } 54 sort(f+1,f+1+m); 55 printf("%d ",bsearch(0,m)); 56 return 0; 57 }