hdu 6181 Two Paths

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.

 

//http://blog.csdn.net/qq_37025443/article/details/77541137  这里有一些讲解，不钻牛角尖了，会用就行，最近要补题了 

 

就是让求次短路，从网上学来的模板 （加油，能多学就多学嘛，反正又没坏处

 

#include<bits/stdc++.h>
using namespace std;
int n,m;
typedef long long ll;
typedef pair<ll,int> pli;//前dis 后node
ll dis1[100100],dis2[100100];
struct node{
    int to,cost;
    node(int t,int c):to(t),cost(c){}
};
vector<node>G[100100];

void solve()
{
    priority_queue<pli,vector<pli>,greater<pli> > Q;
    dis1[1] =0;
    Q.push({0,1});
    while (!Q.empty())
    {
        pli now = Q.top(); Q.pop();
        int v = now.second;
        ll d1 = now.first;
        if(dis2[v] < d1) //说明这个值 太大 不需要更新了
            continue;
        for(int i=0;i<G[v].size();i++)
        {
            node e = G[v][i];
            int to = e.to;
            ll d2 = d1+e.cost;
            if(d2 < dis1[to])
            {
                swap(d2,dis1[to]);
                Q.push({dis1[to],to});
            }
            if(d2 <dis2[to]&& d2 > dis1[to])
            {
                swap(d2,dis2[to]);
                Q.push({dis2[to],to});
            }
        }
    }
    cout<<dis2[n]<<endl;
}

int main ()
{
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d %d",&n,&m);
        for(int i=0;i<=n;i++)
        {
            dis1[i]=dis2[i]=1e18;
            G[i].clear();
        }
        for(int i=1;i<=m;i++)
        {
            int x,y,cost;
            scanf("%d %d %d",&x,&y,&cost);
            G[x].push_back({y,cost});
            G[y].push_back({x,cost});
        }
        solve();
    }
}