u011134403 于 2013.06.24 14:46 提问

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.

The input includes several cases. For each case, the first line contains the number of farms, N (3 <= N <= 500). The following lines contain the N x N conectivity matrix, where each element shows the distance from one farm to another. Logically, they are N lines of N space-separated integers. Of course, the diagonal will be 0, since the distance from farm i to itself is not interesting for this problem

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.

6
0 6 1 5 100 100
6 0 5 100 3 100
1 5 0 5 6 4
5 100 5 0 100 2
100 3 6 100 0 6
100 100 4 2 6 0

15

optical fiber 　　光缆

N x N conectivity matrix N x N的邻接矩阵

space-separated integers　　空格隔开的整数

the diagonal　对角线

1个回答

u014296677   2014.06.13 10:45

#include

#define Max 200000
#define N 1100
int a[N][N],low[N],visited[N];
int n;

int prim()

{
int i,j,t,Min,sum=0;
visited[1]=1;t=1;
for(i=1;i<=n;i++)
if(i!=t) low[i]=a[t][i];
for(i=1;i {
Min=Max;
for(j=1;j if(visited[j]==0&&Min>low[j])
{
Min=low[j];t=j;
}
sum+=Min;
visited[t]=1;
for(j=1;j<=n;j++)
if(visited[j]==0&&low[j]>a[t][j])
low[j]=a[t][j];
}
return sum;
}

int main()
{
int i,j,k;
while(scanf("%d",&n)!=EOF)
{
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
{
scanf("%d",&a[i][j]);
}
k=prim();
printf("%d\n",k);
}
return 0;
}