Description
Farmer John has grown so lazy that he no longer wants to continue
maintaining the cow paths that currently provide a way to visit
each of his N (5 <= N <= 10,000) pastures (conveniently numbered
1..N). Each and every pasture is home to one cow. FJ plans to remove
as many of the P (N-1 <= P <= 100,000) paths as possible while keeping
the pastures connected. You must determine which N-1 paths to keep.
Bidirectional path j connects pastures S_j and E_j (1 <= S_j <= N;
1 <= E_j <= N; S_j != E_j) and requires L_j (0 <= L_j <= 1,000) time
to traverse. No pair of pastures is directly connected by more than
one path.
The cows are sad that their transportation system is being reduced.
You must visit each cow at least once every day to cheer her up.
Every time you visit pasture i (even if you're just traveling
through), you must talk to the cow for time C_i (1 <= C_i <= 1,000).
You will spend each night in the same pasture (which you will choose)
until the cows have recovered from their sadness. You will end up
talking to the cow in the sleeping pasture at least in the morning
when you wake up and in the evening after you have returned to
sleep.
Assuming that Farmer John follows your suggestions of which paths
to keep and you pick the optimal pasture to sleep in, determine the
minimal amount of time it will take you to visit each cow at least
once in a day.
Input
* Line 1: Two space-separated integers: N and P
* Lines 2..N+1: Line i+1 contains a single integer: C_i
* Lines N+2..N+P+1: Line N+j+1 contains three space-separated
integers: S_j, E_j, and L_j
Output
* Line 1: A single integer, the total time it takes to visit all the
cows (including the two visits to the cow in your
sleeping-pasture)
Sample Input
Sample Output
HINT
INPUT DETAILS:
OUTPUT DETAILS:
Keep these paths:
Wake up in pasture 4 and visit pastures in the order 4, 5, 4, 2,
3, 2, 1, 2, 4 yielding a total time of 176 before going back to
sleep.
#include<stdio.h>
#include<algorithm>
using namespace std;
long long min(long long x,long long y)
{
if(x>y)
return y;
return x;
}
int a[10001];
struct Edge
{
int x,y,val;
}edge[200001];
int fa[10001];
int find(int x)
{
if(fa[x]!=x)
fa[x]=find(fa[x]);
return fa[x];
}
bool cmp(const Edge &a,const Edge &b)
{
return a.val<b.val;
}
long long ans=1e17;
int main()
{
int n,m,i;
scanf("%d%d",&n,&m);
for(i=1;i<=n;i++)
{
scanf("%d",&a[i]);
ans=min(ans,a[i]);
fa[i]=i;
}
for(i=1;i<=m;i++)
{
scanf("%d%d%d",&edge[i].x,&edge[i].y,&edge[i].val);
edge[i].val=2*edge[i].val+a[edge[i].x]+a[edge[i].y];
}
sort(edge+1,edge+m+1,cmp);
for(i=1;i<=m;i++)
{
int fx=find(edge[i].x);
int fy=find(edge[i].y);
if(fx!=fy)
{
ans+=edge[i].val;
fa[fx]=fy;
}
}
printf("%d",ans);
}