HDU-4725 The Shortest Path in Nya Graph 最短路

时间：2022-09-21 15:53:41

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4725

　　如果直接建图复杂度过大，但是考虑到每层之间的有效边很少，只要在每层增加两个虚拟节点n+i和2*n+i。n+i节点向 i 层的所有连边，权值为0。i 层的所有点向2*n+i节点连边，权值为0。然后每层直接建立边就可以了，即2*n+i-1向n+i连边，权值为c，2*n+i向n+i-1连边，权值为c。3*n个点，最多有有9*n条边。。

 //STATUS:C++_AC_730MS_14340KB

 #include <functional>

 #include <algorithm>

 #include <iostream>

 //#include <ext/rope>

 #include <fstream>

 #include <sstream>

 #include <iomanip>

 #include <numeric>

 #include <cstring>

 #include <cassert>

 #include <cstdio>

 #include <string>

 #include <vector>

 #include <bitset>

 #include <queue>

 #include <stack>

 #include <cmath>

 #include <ctime>

 #include <list>

 #include <set>

 #include <map>

 using namespace std;

 //#pragma comment(linker,"/STACK:102400000,102400000")

 //using namespace __gnu_cxx;

 //define

 #define pii pair<int,int>

 #define mem(a,b) memset(a,b,sizeof(a))

 #define lson l,mid,rt<<1

 #define rson mid+1,r,rt<<1|1

 #define PI acos(-1.0)

 //typedef

 typedef __int64 LL;

 typedef unsigned __int64 ULL;

 //const

 const int N=;

 const int INF=0x3f3f3f3f;

 const int MOD=,STA=;

 const LL LNF=1LL<<;

 const double EPS=1e-;

 const double OO=1e60;

 const int dx[]={-,,,};

 const int dy[]={,,,-};

 const int day[]={,,,,,,,,,,,,};

 //Daily Use ...

 inline int sign(double x){return (x>EPS)-(x<-EPS);}

 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}

 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}

 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}

 template<class T> inline T Min(T a,T b){return a<b?a:b;}

 template<class T> inline T Max(T a,T b){return a>b?a:b;}

 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}

 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}

 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}

 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}

 //End

 struct Edge{

     int u,v,w;

 }e[*N];

 int first[N],next[*N];

 LL d[N];

 int S,T,n,m,c,mt;

 void adde(int a,int b,int c)

 {

     e[mt].u=a,e[mt].v=b;e[mt].w=c;

     next[mt]=first[a],first[a]=mt++;

 }

 #define pli pair<LL,int>

 LL dijkstra(int s)

 {

     int i,j,u,v,x;

     pli t;

     priority_queue<pli,vector<pli>,greater<pli> > q;

     for(i=;i<=*n;i++)d[i]=LNF;

     d[s]=;

     q.push(make_pair(d[s],s));

     while(!q.empty()){

         t=q.top();q.pop();

         u=t.second;

         if(t.first!=d[u])continue;

         for(i=first[u];i!=-;i=next[i]){

             if(d[u]+e[i].w<d[e[i].v]){

                 d[e[i].v]=d[u]+e[i].w;

                 q.push(make_pair(d[e[i].v],e[i].v));

             }

         }

     }

     return d[T];

 }

 int main(){

 //   freopen("in.txt","r",stdin);

     int Ca,i,j,k,a,b,w,ca=;

     scanf("%d",&Ca);

     while(Ca--)

     {

         scanf("%d%d%d",&n,&m,&c);

         S=,T=n;

         mem(first,-);mt=;

         for(i=;i<=n;i++){

             scanf("%d",&a);

             adde(n+a,i,);

             adde(i,(n<<)+a,);

         }

         for(i=;i<=n;i++){

             adde((n<<)+i-,n+i,c);

             adde((n<<)+i,n+i-,c);

         }

         for(i=;i<m;i++){

             scanf("%d%d%d",&a,&b,&w);

             adde(a,b,w);

             adde(b,a,w);

         }

         dijkstra(S);

         if(d[T]==LNF)d[T]=-;

         printf("Case #%d: %I64d\n",ca++,d[T]);

     }

     return ;

 }

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