HDU 6181 Two Paths

Two Paths

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 153428/153428 K (Java/Others)
Total Submission(s): 110    Accepted Submission(s): 70

[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

 

[align=left]Source[/align]
2017 Multi-University Training Contest - Team 10
 

[align=left]Recommend[/align]
liuyiding   |   We have carefully selected several similar problems for you:  6181 6180 6179 6178 6177 

题意：
求出与最短路不相同的一条次短路(似乎并不是严格的次短路，只需稍微与最短路有点不同即可)
思路：
我直接套上第k短路板子过了，因为是无向图，所以不用再建新图。
第k短路完整版的板子：http://blog.csdn.net/z_mendez/article/details/47057461
示例程序：

#include <bits/stdc++.h>
#define MAX 100000000000000
using namespace std;
struct jj
{
int v,next;
long long c;
}w[200000];
struct kk
{
int v;
long long g,f;
bool operator<(const struct kk &a) const
{
if(a.f<f)
{
return 1;
}
else if(a.f==f&&a.g<g)
{
return 1;
}
else
{
return 0;
}
}
};
int h[100001],numw;
long long dis[100001];
void insert(int u,int v,long long c)
{
w[numw].v=v;
w[numw].c=c;
w[numw].next=h[u];
h[u]=numw++;
}
void spfa(int s)
{
int i,v[100001];
queue<int>q;
for(i=1;s>=i;i++)
{
dis[i]=MAX;
v[i]=0;
}
dis[s]=0;
q.push(s);
while(q.empty()==0)
{
s=q.front();
q.pop();
v[s]=0;
for(i=h[s];i!=-1;i=w[i].next)
{
if(dis[w[i].v]>dis[s]+w[i].c)
{
dis[w[i].v]=dis[s]+w[i].c;
if(v[w[i].v]==0)
{
v[w[i].v]=1;
q.push(w[i].v);
}
}
}
}
}
long long A_star(int s,int t)
{
int i,v[100001],num=0;
struct kk pre,pos;
priority_queue<struct kk>q;
memset(v,0,sizeof(v));
pos.v=s;
pos.g=0;
pos.f=pos.g+dis[s];
q.push(pos);
while(q.empty()==0)
{
pre=q.top();
q.pop();
for(i=h[pre.v];i!=-1;i=w[i].next)
{
if(v[w[i].v]==3)           //剪枝,一个点最多访问三次
{
continue;
}
pos.v=w[i].v;
pos.g=pre.g+w[i].c;
pos.f=pos.g+dis[w[i].v];
if(pos.v==t)            //找到一条最短路径
{
num++;
}
if(num==2)              //找到k(次)短路
{
return pos.g;
}
v[w[i].v]++;
q.push(pos);
}
}
}
int main()
{
int t,n,m,i,u,v;
long long c;
scanf("%d",&t);
while(t--)
{
memset(h,-1,sizeof(h));
numw=0;
scanf("%d %d",&n,&m);
for(i=1;m>=i;i++)
{
scanf("%d %d %lld",&u,&v,&c);
insert(u,v,c);
insert(v,u,c);
}
spfa(n);
printf("%lld\n",A_star(1,n));
}
return 0;
}