HDU 6181 次短路

Two Paths

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 153428/153428 K (Java/Others)
Total Submission(s): 235    Accepted Submission(s): 137

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：给出一个图，要求找出不是最短情况下的最短路，也就是次短路。

用两个数组, dis1[]  和 dis2[] dis1记录最短路，dis2记录次短路，在Dijkstra 记录最短路的同时，只要计算的距离

满足  dis2[v] > dis > dis1[v] ，那就把 dis2[v] 更新为 dis ，就这样一遍计算最短路一遍记录次短路就ok

#include<cstdio>#include<iostream>#include<algorithm>#include<string.h>#include<queue>using namespace std;#define ll long long#define maxn 100005typedef pair<ll,ll> p;const ll inf = 1000000000100100;struct node{ll to,val;node(ll x,ll y):to(x),val(y) {}};vector<node>e[maxn];vector<node>len;ll n,m,dis1[maxn],dis2[maxn];void dijkstra(){priority_queue<p,vector<p>,greater<p> >q;for(int i = 1;i <= n;i++){dis1[i] = dis2[i] = inf;}dis1[1] = 0;q.push(p(dis1[1],1));while(!q.empty()){ll now = q.top().first;ll to = q.top().second;//printf("q top = %lld %lld\n",q.top().first,q.top().second);q.pop();if(dis2[to] < now)continue;for(int i = 0;i < e[to].size();i++){ll t = e[to][i].to;ll v = e[to][i].val;ll len = now + v;if(dis1[t] > len){swap(len,dis1[t]);q.push(p(dis1[t],t));}if(dis2[t] > len && dis1[t] < len){dis2[t] = len;q.push(p(dis2[t],t));}}}//printf("%lld\n",dis1[n]);printf("%lld\n",dis2[n]);}void add(ll u,ll v,ll w){e[u].push_back(node(v,w));e[v].push_back(node(u,w));}void init(){for(int i = 0;i < maxn;i++){e[i].clear();}len.clear();}int main() {int t;ll u,v,w;scanf("%d",&t);while(t--){scanf("%d %d",&n,&m);init();for(int i = 1;i <= m;i++){scanf("%lld %lld %lld",&u,&v,&w);add(u,v,w);}dijkstra();}return 0;}