hdu 6181 (次短路

传送门

Two Paths

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 153428/153428 K (Java/Others)
Total Submission(s): 321    Accepted Submission(s): 184

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

23 31 2 12 3 41 3 32 11 2 1

 

Sample Output

53
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

 

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次短路不断更新最短的路径和次短的路径，到最后一定是次短路径加最短路径。

/*裸的次短路不断更新v->u的次短路，直到v->u的次短路只比最短路小*/#include <vector>#include <iostream>#include <stdio.h>#include <queue>using namespace std;#define ll long long#define INF 1e18#define MAXM 100100struct edge{int to;ll w;};typedef pair<ll,int>P;int n,r;ll d[MAXM];ll d1[MAXM];vector<edge>G[MAXM];bool vis[MAXM];void solve(){    priority_queue<P, vector<P>,greater<P> >que;    fill(d+1,d+n+1,INF);    fill(d1+1,d1+n+1,INF);    d[1]=0;    que.push(P(0,1));    while(!que.empty())    {        P p=que.top();        que.pop();        int v=p.second;        if(d1[v]<p.first)continue;        for(int i=0;i<G[v].size();i++)        {            edge &e=G[v][i];            ll d2=p.first+e.w;            if(d[e.to]>d2)            {                swap(d[e.to],d2);                que.push(P(d[e.to],e.to));            }            if(d1[e.to]>d2&&d[e.to]<d2)            {                d1[e.to]=d2;                que.push(P(d1[e.to],e.to));            }        }    }    printf("%I64d\n",d1[n]);}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d %d",&n,&r);        for(int i=1;i<=n;i++)            G[i].clear();        int u,v;        ll w;        while(r--)        {            scanf("%d %d %I64d",&u,&v,&w);            G[u].push_back(edge{v,w});            G[v].push_back(edge{u,w});        }        solve();    }    return 0;}