HDU

Two Paths

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

 

题意：求真 · 次短路。一条路可以走多遍，如果最短路有多种，次短路其实就是另外一种最短路……WA的应该都是这里吧……还有就是INF不能用int的，要用LL的（其实应该是可以int的吧）……我感受到了出题者满满的恶意。

解题思路：套个次短路模板就好了，然后将里面的<号改成<=号。用SPFA会超时不知道为什么……还是我写搓了，改成了迪杰斯特拉就过了。思路其实就是，存最短路的同时存下次短路 ……

#include<iostream>#include<deque>#include<memory.h>#include<stdio.h>#include<map>#include<string>#include<algorithm>#include<vector>#include<math.h>#include<stack>#include<queue>#include<set>#define INF (1LL<<62)//WA了好几发#define ll long long intusing namespace std;typedef pair<ll,int>P;struct edge{    int to;    ll dis;    edge(int to,ll dis){        this -> to = to;        this -> dis = dis;    }};int N,R;int a,b;ll c;ll dis[100005];         //记录最短路径ll disc[100005];       //记录次短路径vector<edge>G[100005];void dijkstra(){    fill(dis,dis+100005,INF);    fill(disc,disc+100005,INF);    priority_queue<P,vector<P>,greater<P> >q;    dis[1]=0;    q.push(P(0,1));    while(q.size()){        P p=q.top();        q.pop();        ll dd=p.first;        int v=p.second;        if(disc[v]<dd) continue;        for(int i=0;i<G[v].size();i++){            edge& e=G[v][i];            ll d=dd+e.dis;            if(dis[e.to]>=d){                ll ttt=d;                d=dis[e.to];                dis[e.to]=ttt;                q.push(P(dis[e.to],e.to));            }            if(disc[e.to]>=d&&dis[e.to]<=d){                disc[e.to]=d;                q.push(P(disc[e.to],e.to));            }        }    }    cout<<disc[N]<<endl;}int main(){    int t;    scanf("%d",&t);    while(t--){        for(int i=0;i<100005;i++)            G[i].clear();    scanf("%d%d",&N,&R);    for(int i=1;i<=R;i++){        scanf("%d%d%lld",&a,&b,&c);        G[a].push_back(edge(b,c));        G[b].push_back(edge(a,c));    }    dijkstra();    }    return 0;}