hdu 6181 Two Paths(次短路)

Two Paths

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 153428/153428 K (Java/Others)
Total Submission(s): 150    Accepted Submission(s): 88

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:

在图中找一条从1->n的次短路，可以走回头路

解析：

次短路模板，类似poj3255

#include<cstdio>#include<cstring>#include<queue>#include<functional>#include<vector>#include<algorithm>using namespace std;typedef long long int ll;#define MAXM  350010#define MAXN 250010#define INF 1223372036854775807typedef pair<ll,int> pp;typedef struct node{int v;ll w;node(int tv=0,ll tw=0):v(tv),w(tw){};}node;int n,m;vector<node> edge[MAXM];ll dis[MAXN],dis2[MAXN];   //dis[i]表示最短路，dis2[i]表示次短路void solve(){fill(dis+1,dis+n+1,INF);fill(dis2+1,dis2+n+1,INF);    priority_queue<pp, vector<pp>, greater<pp> > q;  //用优先队列加速搜索dis[1]=0;q.push(pp(dis[1],1));   //second是该边指向（虽然是无向的）的顶点，first是这条边的权值while(q.size()){pp p=q.top(); q.pop();int v=p.second;ll d=p.first;if(dis2[v]<d) continue;  //如果当前取出的值不是到v的最短路或次短路就contniue,因为v->e的最短边和次短边一定是由->v的最短边和次短边+edge(v,e)得到for(int i=0;i<edge[v].size();i++){int e=edge[v][i].v;ll d2=d+edge[v][i].w;if(dis[e]>d2){swap(dis[e],d2);q.push(pp(dis[e],e));}if(dis2[e]>d2&&d2>dis[v])  //d2>dis[v]防止d2小于dis[v],这样v->e就变成负权边了，但可有可无{dis2[e]=d2;q.push(pp(dis2[e],e));}}}printf("%lld\n",dis2[n]);}int main(){int t;scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);for(int i=1;i<=n;i++) edge[i].clear();for(int i=0;i<m;i++){int a,b;ll c;scanf("%d%d%lld",&a,&b,&c);edge[a].push_back(node(b,c));edge[b].push_back(node(a,c));}solve();}}