1170: Wire Is So Expensive

1170: Wire Is So Expensive

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Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	620	259	Standard

1st Jilin University ACM International Collegiate Programming Contest

Assume you are working for ACM(Andrew Communication Management) company. This company is specialized to make towns connected by wire. We all know that wire can only be put along roads. But in recent years, the price of per metre of wire is going higher and higher. So the company needs to know what is the minimal length of wire they have to prepare to connect several towns together in order that each town can connect to another by wire. You are given a map of N towns and how they are connected by roads. Your task is to write a program to determine the minimal length of wire.

Input Specification

This problem contains M maps. The first line of input is an integer M. Then follow M maps' descriptions, each of which is described below:

N
a1 b1 c1
a2 b2 c2
...
an bn cn
0 0 0

where N(1<=N<=20) is the number of towns(1, 2, ... N) and ai, bi, ci means there is a road between towns ai and bi(1<=ai, bi<=N) with length ci(1<=ci<=100). Three 0s mark the end of a map. All the numbers will be integers.

Output Specification

For each map, you should print a line containing the minimal length of wire the company has to prepare.

Sample Input

1
5
1 2 3
1 4 5
2 3 6
2 4 2
2 5 6
3 5 5
4 5 8
0 0 0

Sample Output

16

#include<stdio.h>
#include<string.h>
int inf=(1<<31)-2;
int main()
{
    int c[101][101];
    int ll;
    scanf("%d",&ll);
    while(ll--)
    {
        int n,i,j;
        scanf("%d",&n);
        int ai,bi,ci;
        memset(c,12,sizeof(c));
        while(1)
        {
            scanf("%d%d%d",&ai,&bi,&ci);
            if(ai==0&&bi==0&&ci==0) break;
            if(ci<c[ai-1][bi-1]) c[ai-1][bi-1]=c[bi-1][ai-1]=ci;
        }
        int vis[100],lowc[100];
        memset(vis,0,sizeof(vis));
        int minc,res=0;
        int p=0;
        vis[0]=1;
        for(i=0;i<n;i++) lowc[i]=c[0][i];
        for(j=1;j<n;j++)
        {
            minc=inf;p=0;
            for(i=0;i<n;i++)
            {
                if(vis[i]==0&&minc>lowc[i])
                {
                    minc=lowc[i];
                    p=i;
                }
            }
            if(minc==inf) break;
            vis[p]=1;res+=minc;
            for(i=0;i<n;i++)
            {
                if(vis[i]==0&&lowc[i]>c[p][i])  lowc[i]=c[p][i];
            }
        }
        printf("%d/n",res);
    }
    return 0;
}