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  • HDU 6181 Two Paths【次短路】【模板题】

    Two Paths

    Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 153428/153428 K (Java/Others)
    Total Submission(s): 236    Accepted Submission(s): 138


    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

    套一个次短路模板即可

    #include <bits/stdc++.h>
    #define INF 1e16+100
    #define ms(x,y) memset(x,y,sizeof(x))
    using namespace std;
    
    typedef long long ll;
    typedef pair<ll,ll> P;
    
    const double pi = acos(-1.0);
    const int mod = 1e9 + 7;
    const int maxn = 1e5 + 5;
    
    struct Edge{
    	ll to,cost;
    };
    
    ll n,m;
    vector<Edge> a[maxn];
    ll dist[maxn],dist2[maxn];
    
    void addedge(ll u,ll v,ll w)
    {
    	a[u].push_back(Edge{v,w});
    	a[v].push_back(Edge{u,w});
    }
    
    void solve()
    {
    	priority_queue<P, vector<P>, greater<P> >que;
    	//ms(dist,INF);
    	//ms(dist2,INF);
    	fill(dist,dist+n,INF);
    	fill(dist2,dist2+n,INF);
    	dist[0]=0;
    	que.push(P(0,0));
    	while(que.size())
    	{
    		P u=que.top();que.pop();
    		int v=u.second;
    		ll d=u.first;
    		if(dist2[v]<d) continue;	//不是次短距离则抛弃
    		for(int i=0;i<a[v].size();i++)
    		{
    			Edge e=a[v][i];
    			ll d2=d+e.cost;
    			if(dist[e.to]>d2)	//更新最短
    			{
    				swap(dist[e.to],d2);
    				que.push(P(dist[e.to],e.to));
    			}
    			if(dist2[e.to]>d2&&dist[e.to]<d2)	//更新次短
    			{
    				dist2[e.to]=d2;
    				que.push(P(dist2[e.to],e.to));
    			}
    		}
    	}
    	printf("%lld
    ",dist2[n-1]);
    }
    
    int main()
    {
    	//freopen("in.txt","r",stdin);
    	//freopen("out.txt","w",stdout);
    	int t;
    	scanf("%d",&t);
    	while(t--)
    	{
    		scanf("%lld%lld",&n,&m);
    		for(int i=0;i<n;i++) a[i].clear();
    		for(int i=0;i<m;i++)
    		{
    			ll p,q,w;
    			scanf("%lld%lld%lld",&p,&q,&w);
    			addedge(p-1,q-1,w);
    		}
    		solve();
    	}
    	return 0;
    }

    Fighting~
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  • 原文地址:https://www.cnblogs.com/Archger/p/8451561.html
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