Two Paths

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 153428/153428 K (Java/Others)

Total Submission(s): 236    Accepted Submission(s): 138

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

套一个次短路模板即可

#include 
    
     #define INF 1e16+100#define ms(x,y) memset(x,y,sizeof(x))using namespace std;typedef long long ll;typedef pair
     
       P;const double pi = acos(-1.0);const int mod = 1e9 + 7;const int maxn = 1e5 + 5;struct Edge{	ll to,cost;};ll n,m;vector
      
        a[maxn];ll dist[maxn],dist2[maxn];void addedge(ll u,ll v,ll w){	a[u].push_back(Edge{v,w});	a[v].push_back(Edge{u,w});}void solve(){	priority_queue
       
        , greater
        
 >que;	//ms(dist,INF);	//ms(dist2,INF);	fill(dist,dist+n,INF);	fill(dist2,dist2+n,INF);	dist[0]=0;	que.push(P(0,0));	while(que.size())	{		P u=que.top();que.pop();		int v=u.second;		ll d=u.first;		if(dist2[v]
         
          d2) //更新最短 { swap(dist[e.to],d2); que.push(P(dist[e.to],e.to)); } if(dist2[e.to]>d2&&dist[e.to]