HNU ACM: Cheering up the Cows (kruscal)

B: Cheering up the Cows

Time Limit: 1000 ms     Case Time Limit: 1000 ms     Memory Limit: 65536 KB
Submit: 13     Accepted: 8

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

 

5 7
10
10
20
6
30
1 2 5
2 3 5
2 4 12
3 4 17
2 5 15
3 5 6
4 5 12

Sample Output

 

176

Hint

INPUT DETAILS:

             +-(15)-+
            /        /
           /          /
    1-(5)-2-(5)-3-(6)--5
           /   /(17)  /
        (12)/ /      /(12)
             4------+

OUTPUT DETAILS:

Keep these paths:

    1-(5)-2-(5)-3      5
           /          /
        (12)/        /(12)
            *4------+

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 <string.h>
#include "algorithm"
#define maxe 100100
#define maxn 10100
using namespace std;
struct edge
{
int e,v;
int weight;
}path[maxe];
bool operator <(edge a,edge b)
{
return a.weight<b.weight;
}
int w[maxn];
int p[maxn],rank[maxn];
int n,m;
int find(int t)
{
if(t==p[t]) return p[t];
p[t]=find(p[t]);
return p[t];
}
void union_set(int x,int y)
{
int s=find(x),t=find(y);
if(rank[s]>rank[t]) p[t]=s;
else
{
p[s]=t;
if(rank[s]==rank[t]) rank[t]++;
}
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
int small=0x3fffffff;
for(int i = 1;i<=n;i++)
{
scanf("%d",&w[i]);
if(small>w[i]) small=w[i];
p[i]=i;rank[i]=0;
}
for(int i=0;i<m;i++)
{
int x,y,z;
scanf("%d%d%d",&x,&y,&z);
path[i].e=x;path[i].v=y;
path[i].weight=z*2+w[x]+w[y];
}
sort(path,path+m);
int ans=0;int cnt=0;
for(int i=0;i<m;i++)
{
if(find(path[i].e)!=find(path[i].v))
{
ans+=path[i].weight;
cnt++;
union_set(path[i].e,path[i].v);
}
if(cnt==n-1) break;
}
printf("%d/n",ans+small);
}
return 0;
}