HDU 6181 第k短路

Two Paths

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 153428/153428 K (Java/Others)
Total Submission(s): 525    Accepted Submission(s): 265

Problem Description

You are given a undirected graph with n nodes (numbered from 1 to n) and m edges. Alice and Bob are now trying to play a game. 
Both of them will take different route from 1 to n (not necessary simple).
Alice always moves first and she is so clever that take one of the shortest path from 1 to n.
Now is the Bob's turn. Help Bob to take possible shortest route from 1 to n.
There's neither multiple edges nor self-loops.
Two paths S and T are considered different if and only if there is an integer i, so that the i-th edge of S is not the same as the i-th edge of T or one of them doesn't exist.

 

Input

The first line of input contains an integer T(1 <= T <= 15), the number of test cases.
The first line of each test case contains 2 integers n, m (2 <= n, m <= 100000), number of nodes and number of edges. Each of the next m lines contains 3 integers a, b, w (1 <= a, b <= n, 1 <= w <= 1000000000), this means that there's an edge between node a and node b and its length is w.
It is guaranteed that there is at least one path from 1 to n.
Sum of n over all test cases is less than 250000 and sum of m over all test cases is less than 350000.

 

Output

For each test case print length of valid shortest path in one line.

 

Sample Input

2
3 3
1 2 1
2 3 4
1 3 3
2 1
1 2 1

 

Sample Output

5
3

Hint

For testcase 1, Alice take path 1 - 3 and its length is 3, and then Bob will take path 1 - 2 - 3 and its length is 5.
For testcase 2, Bob will take route 1 - 2 - 1 - 2 and its length is 3

 

Source

2017 Multi-University Training Contest - Team 10

 

题解：无向图求第k短路（非次短路） 模板

 #include<iostream>
 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 #include<map>
 #include<queue>
 #include<stack>
 #include<vector>
 #include<set>
 using namespace std;
 typedef long long ll;
 typedef pair<ll,int> P;
 const int maxn=2e5+,maxm=2e5+,inf=0x3f3f3f3f,mod=1e9+;
 const ll INF=1e17+;
 struct edge
 {
     int from,to;
     ll w;
 } pre[maxn];
 vector<edge>G[maxn],T[maxn];
 priority_queue<P,vector<P>,greater<P> >q;
 ll dist[maxn];

 void addedge(int u,int v,ll w)
 {
     G[u].push_back((edge)
     {
         u,v,w
     });
     T[v].push_back((edge)
     {
         v,u,w
     });
 }
 void dij(int s)
 {
     dist[s]=0LL;
     q.push(P(dist[s],s));
     while(!q.empty())
     {
         P p=q.top();
         q.pop();
         int u=p.second;
         for(int i=; i<T[u].size(); i++)
         {
             edge e=T[u][i];
             if(dist[e.to]>dist[u]+e.w)
             {
                 dist[e.to]=dist[u]+e.w;
                 q.push(P(dist[e.to],e.to));
             }
         }
     }
 }
 struct node
 {
     int to;
     ///g(p)为当前从s到p所走的路径的长度;dist[p]为点p到t的最短路的长度;
     ll g,f;///f=g+dist,f(p)的意义为从s按照当前路径走到p后再走到终点t一共至少要走多远;
     bool operator<(const node &x ) const
     {
         if(x.f==f) return x.g<g;
         return x.f<f;
     }
 };

 ll A_star(int s,int t,int k)
 {
     priority_queue<node>Q;
     if(dist[s]==INF) return -;
     int cnt=;
     if(s==t) k++;
     ll g=0LL;
     ll f=g+dist[s];
     Q.push((node)
     {
         s, g, f
     });
     while(!Q.empty())
     {
         node x=Q.top();
         Q.pop();
         int u=x.to;
         if(u==t) cnt++;
         if(cnt==k) return x.g;
         for(int i=; i<G[u].size(); i++)
         {
             edge e=G[u][i];
             ll g=x.g+e.w;
             ll f=g+dist[e.to];
             Q.push((node)
             {
                 e.to, g, f
             });
         }
     }
     return -;
 }
 void init(int n)
 {
     for(int i=; i<=n+; i++) G[i].clear(),T[i].clear();
 }
 int main()
 {
     int n,m;
     int t;
     scanf("%d",&t);
     while(t--){
     scanf("%d%d",&n,&m);
     for(int i=; i<=m; i++)
     {
         int u,v;
         ll w;
         scanf("%d%d%lld",&u,&v,&w);
         addedge(u,v,w);
         addedge(v,u,w);
     }
     for(int i=; i<=n; i++) dist[i]=INF;
     dij(n);
     printf("%lld\n",A_star(,n,));
     init(n);
     }
     return ;
 }