HDU - 6181 Two Paths（真 · 求次短路径）

Two Paths

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

 

题意：求真 · 次短路。一条路可以走多遍，如果最短路有多种，次短路其实就是另外一种最短路……WA的应该都是这里吧……还有就是INF不能用int的，要用LL的（其实应该是可以int的吧）……我感受到了出题者满满的恶意。

解题思路：套个次短路模板就好了，然后将里面的<号改成<=号。用SPFA会超时不知道为什么……还是我写搓了，改成了迪杰斯特拉就过了。思路其实就是，存最短路的同时存下次短路 ……

#include<iostream>
#include<deque>
#include<memory.h>
#include<stdio.h>
#include<map>
#include<string>
#include<algorithm>
#include<vector>
#include<math.h>
#include<stack>
#include<queue>
#include<set>
#define INF (1LL<<62)//WA了好几发
#define ll long long int
using namespace std;

typedef pair<ll,int>P;
struct edge{
int to;
ll dis;
edge(int to,ll dis){
this -> to = to;
this -> dis = dis;
}
};
int N,R;
int a,b;
ll c;
ll dis[100005];         //记录最短路径
ll disc[100005];       //记录次短路径
vector<edge>G[100005];

void dijkstra(){
fill(dis,dis+100005,INF);
fill(disc,disc+100005,INF);
priority_queue<P,vector<P>,greater<P> >q;
dis[1]=0;
q.push(P(0,1));
while(q.size()){
P p=q.top();
q.pop();
ll dd=p.first;
int v=p.second;
if(disc[v]<dd) continue;
for(int i=0;i<G[v].size();i++){
edge& e=G[v][i];
ll d=dd+e.dis;
if(dis[e.to]>=d){
ll ttt=d;
d=dis[e.to];
dis[e.to]=ttt;
q.push(P(dis[e.to],e.to));
}
if(disc[e.to]>=d&&dis[e.to]<=d){
disc[e.to]=d;
q.push(P(disc[e.to],e.to));
}
}
}
cout<<disc
<<endl;
}

int main()
{
int t;
scanf("%d",&t);
while(t--){

for(int i=0;i<100005;i++)
G[i].clear();

scanf("%d%d",&N,&R);
for(int i=1;i<=R;i++){
scanf("%d%d%lld",&a,&b,&c);
G[a].push_back(edge(b,c));
G[b].push_back(edge(a,c));
}
dijkstra();
}
return 0;
}