HDU 6181 第k短路

Two Paths

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 153428/153428 K (Java/Others)
Total Submission(s): 525    Accepted Submission(s): 265

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

2017 Multi-University Training Contest - Team 10

 

题解：无向图求第k短路（非次短路） 模板

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<set>
using namespace std;
typedef long long ll;
typedef pair<ll,int> P;
const int maxn=2e5+100,maxm=2e5+100,inf=0x3f3f3f3f,mod=1e9+7;
const ll INF=1e17+7;
struct edge
{
    int from,to;
    ll w;
} pre[maxn];
vector<edge>G[maxn],T[maxn];
priority_queue<P,vector<P>,greater<P> >q;
ll dist[maxn];

void addedge(int u,int v,ll w)
{
    G[u].push_back((edge)
    {
        u,v,w
    });
    T[v].push_back((edge)
    {
        v,u,w
    });
}
void dij(int s)
{
    dist[s]=0LL;
    q.push(P(dist[s],s));
    while(!q.empty())
    {
        P p=q.top();
        q.pop();
        int u=p.second;
        for(int i=0; i<T[u].size(); i++)
        {
            edge e=T[u][i];
            if(dist[e.to]>dist[u]+e.w)
            {
                dist[e.to]=dist[u]+e.w;
                q.push(P(dist[e.to],e.to));
            }
        }
    }
}
struct node
{
    int to;
    ///g(p)为当前从s到p所走的路径的长度;dist[p]为点p到t的最短路的长度;
    ll g,f;///f=g+dist,f(p)的意义为从s按照当前路径走到p后再走到终点t一共至少要走多远;
    bool operator<(const node &x ) const
    {
        if(x.f==f) return x.g<g;
        return x.f<f;
    }
};

ll A_star(int s,int t,int k)
{
    priority_queue<node>Q;
    if(dist[s]==INF) return -1;
    int cnt=0;
    if(s==t) k++;
    ll g=0LL;
    ll f=g+dist[s];
    Q.push((node)
    {
        s, g, f
    });
    while(!Q.empty())
    {
        node x=Q.top();
        Q.pop();
        int u=x.to;
        if(u==t) cnt++;
        if(cnt==k) return x.g;
        for(int i=0; i<G[u].size(); i++)
        {
            edge e=G[u][i];
            ll g=x.g+e.w;
            ll f=g+dist[e.to];
            Q.push((node)
            {
                e.to, g, f
            });
        }
    }
    return -1;
}
void init(int n)
{
    for(int i=0; i<=n+10; i++) G[i].clear(),T[i].clear();
}
int main()
{
    int n,m;
    int t;
    scanf("%d",&t);
    while(t--){
    scanf("%d%d",&n,&m);
    for(int i=1; i<=m; i++)
    {
        int u,v;
        ll w;
        scanf("%d%d%lld",&u,&v,&w);
        addedge(u,v,w);
        addedge(v,u,w);
    }
    for(int i=0; i<=n; i++) dist[i]=INF;
    dij(n);
    printf("%lld
",A_star(1,n,2));
    init(n);
    }
    return 0;
}