2017多校联合第四场1011/hdu6181（次短路）

Two Paths

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 153428/153428 K (Java/Others)
Total Submission(s): 933    Accepted Submission(s): 431

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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Statistic | Submit | Discuss | Note就是次短路，模板套一下就过了

#include <iostream>#include <cstdio>#include <cstring>#include <vector>#include <queue>#define ll long long#define INF 1000000000000000000using namespace std;typedef pair<ll,ll> P;struct edge{    ll to,v;    edge(ll to,ll v):to(to),v(v) {}    edge() {}};const int maxn = 1e5+10;//const int maxe = 100005;ll V,E;vector<edge> g[maxn];ll d[maxn],d2[maxn];//最短距离和次短距离void dijkstra(int s){    priority_queue<P,vector<P>,greater<P> > pq;    for(int i=1; i<=V; i++)    {        d[i]=INF;        d2[i]=INF;    }    d[s]=0;    pq.push(P(0,s));    while(pq.size())    {        P nowe=pq.top();        pq.pop();        if(nowe.first>d2[nowe.second]) continue;  //如果这个距离比当前次短路长continue        for(int v=0; v<(int)g[nowe.second].size(); v++)        {            edge nexte=g[nowe.second][v];            ll dis=nowe.first+nexte.v;            if(d[nexte.to]>dis)            {                swap(dis,d[nexte.to]);                pq.push(P(d[nexte.to],nexte.to));            }            if(d2[nexte.to]>dis&&d[nexte.to]<dis)//保证最短路是小于这个次短路的            {                d2[nexte.to]=dis;                pq.push(P(d2[nexte.to],nexte.to));//次短路的点进入pq            }        }    }}int main(){    int s;//起点    int t;    cin>>t;    //cout<<INF<<endl;    while(t--)    {        scanf("%d%d",&V,&E);        {            for(int i=1; i<=V; i++)                g[i].clear();            for(int i=1; i<=E; i++)            {                ll f,t,v;                scanf("%lld %lld %lld",&f,&t,&v);                g[f].push_back(edge(t,v));                g[t].push_back(edge(f,v));            }            s=1;//这题默认起点为1            dijkstra(s);            printf("%lld\n",d2[V]);            //cout<<d2[V]<<endl;        }    }    return 0;}