hdu 6181 Two Paths(次短路)

Two Paths

[b]Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 153428/153428 K (Java/Others)

Total Submission(s): 150    Accepted Submission(s): 88
[/b]

[align=left]Problem Description[/align]
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.

 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
For each test case print length of valid shortest path in one line.
 

[align=left]Sample Input[/align]

2
3 3
1 2 1
2 3 4
1 3 3
2 1
1 2 1

 

[align=left]Sample Output[/align]

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:
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 1223372036854775807
typedef pair<ll,int> pp;

typedef struct node
{
int v;
ll w;
nod
4000
e(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
);

}

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();
}
}