Agri-Net

Time Limit: 1000MS

 

Memory Limit: 10000K

Total Submissions: 44670

 

Accepted: 18268

Description

Farmer John has been elected mayor of his town! One of his campaign promises was to bring internet connectivity to all farms in the area. He needs your help, of course. 
Farmer John ordered a high speed connection for his farm and is going to share his connectivity with the other farmers. To minimize cost, he wants to lay the minimum amount of optical fiber to connect his farm to all the other farms. 
Given a list of how much fiber it takes to connect each pair of farms, you must find the minimum amount of fiber needed to connect them all together. Each farm must connect to some other farm such that a packet can flow from any one farm to any other farm. 
The distance between any two farms will not exceed 100,000. 

Input

The input includes several cases. For each case, the first line contains the number of farms, N (3 <= N <= 100). The following lines contain the N x N conectivity matrix, where each element shows the distance from on farm to another. Logically, they are N lines of N space-separated integers. Physically, they are limited in length to 80 characters, so some lines continue onto others. Of course, the diagonal will be 0, since the distance from farm i to itself is not interesting for this problem.

Output

For each case, output a single integer length that is the sum of the minimum length of fiber required to connect the entire set of farms.

Sample Input

4
0 4 9 21
4 0 8 17
9 8 0 16
21 17 16 0

Sample Output

28
求把所有town连接起来所需最少边权值和，把边权值按从小到大排序，如果当前边权值连接的两边没有关系就把当前边权值相加，如果有关系当前值就没有价值，不用浪费资源联通这两个town。到最后肯定会把所有town连接~

#include
#include

using namespace std;

#define N 1000005

int k, n, a, f[N];

struct node
{
int x, y, w;
}P[N];

int cmp(const void *u, const void *b)
{
node c, d;
c = *(node *)u;
d = *(node *)b;

return c.w-d.w;  

}

void init()
{
for(int i = ; i < N; i++)
f[i] = i;
}
int found(int a)
{
if(f[a] != a)
f[a] = found(f[a]);
return f[a];
}

int main()
{
while(cin >> n)
{
k = ;

    init();

    for(int i = ; i <= n; i++)  
        for(int j = ; j <= n; j++)  
        {  
            cin >> a;  
            P\[k\].w = a;  
            P\[k\].x = i, P\[k\].y = j;  
            k++;  
        }

    qsort(P, k, sizeof(P\[\]), cmp);

    int ans = ;

    for(int i = ; i < k; i++)  
{  
    int nx = found(P\[i\].x), ny = found(P\[i\].y);

    if(nx != ny)  
    {  
        f\[nx\] = ny;  
        ans += P\[i\].w;  
    }  
}  
    //int ans = kuscral(1, n);

    cout << ans << endl;  
}  
return ;  

}