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.
InputThe 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.
OutputFor 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 1Sample 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
此題就是求無向圖的次短路。即除去最短路以外的最短路，只要走的不一樣，可以和最短路一樣長。如果只有一條路，那麼有一段就得走兩遍（就像第二個樣例）；
可以說是個板子題，直接套版子；
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
struct mmp {
int to,cost;
mmp(int a,int b) {
to=a;
cost=b;
}
};
ll d1[100001],d2[100001]; //d1爲從1到i的最短路，d2爲次短路
int m,n;
typedef pair<ll,int> p; //用pair對應長度和節點
vector<mmp> mmap[100001]; //用來儲存每條路，用動態數組和結構體，節省時間和空間
void solve() {
priority_queue<p,vector<p>,greater<p> > q; //用優先隊列
q.push({0,1});
while(!q.empty()) {
p now=q.top();
q.pop();
int too=now.second;
ll d=now.first;
if(d>d2[too]) //若比次短路還長，跳過
continue;
for(int i=0; i<mmap[too].size(); i++) {
ll dd=d+mmap[too][i].cost;
int to=mmap[too][i].to;
if(d1[to]>dd) { //若此條路比最短路短，交換並壓入隊列
swap(dd,d1[to]);
q.push({d1[to],to});
}
if(d2[to]>dd&&dd>d1[to]){ //若此條路比次短路短且比最短路長，與次短路交換位置並壓入隊列
swap(dd,d2[to]);
q.push({d2[to],to});
}
}
}
cout<<d2[n]<<endl;
}
int main() {
int t;
cin>>t;
while(t--) {
cin>>n>>m;
for(int i=1; i<=n; i++) { //將每條路都設置爲無窮大，並清空mmap
d1[i]=d2[i]=1e18;
mmap[i].clear();
}
for(int i=1; i<=m; i++) {
int x,y,c;
cin>>x>>y>>c;
mmap[x].push_back(mmp(y,c));
mmap[y].push_back(mmp(x,c));
}
solve();
}
return 0;
}