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.
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.
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 3HintFor 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\n",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~