HDU-4725 The Shortest Path in Nya Graph 最短路

　　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=300010;
const int INF=0x3f3f3f3f;
const int MOD=1000000007,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e60;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//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[3*N];
int first[N],next[3*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=1;i<=3*n;i++)d[i]=LNF;
    d[s]=0;
    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!=-1;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=1;
    scanf("%d",&Ca);
    while(Ca--)
    {
        scanf("%d%d%d",&n,&m,&c);
        S=1,T=n;
        mem(first,-1);mt=0;
        for(i=1;i<=n;i++){
            scanf("%d",&a);
            adde(n+a,i,0);
            adde(i,(n<<1)+a,0);
        }
        for(i=2;i<=n;i++){
            adde((n<<1)+i-1,n+i,c);
            adde((n<<1)+i,n+i-1,c);
        }
        for(i=0;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]=-1;

        printf("Case #%d: %I64d
",ca++,d[T]);
    }
    return 0;
}