Agri-Net
Time Limit: 1000MS
Memory Limit: 10000K
Total Submissions: 43215
Accepted: 17683
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
Source
USACO 102
代碼：
#include
#include
#define MAX 99999999
int mat[105][105];
int main()
{
int n,i,j,k,sum,vis[1005],a,b,c,dist[105];
while(scanf("%d",&n)!=EOF)
{
if(n==0)break;
memset(vis,0,sizeof(vis));
for(i=0;i<105;i++)
for(j=0;j<105;j++)
mat[i][j]=MAX;
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
{
scanf("%d",&a);
mat[i][j]=a;
}
vis[1]=1;
dist[1]=0;
sum=0;
int pos=1;
for(i=2;i<=n;i++)
{
dist[i]=mat[1][i];
}
for(i=1;i {
int mini=MAX,u=-1;
for(j=1;j<=n;j++)
if(!vis[j]&&dist[j] {
u=j;mini=dist[j];
}
vis[u]=1;
sum=sum+dist[u];
for(j=1;j<=n;j++)
if(!vis[j]&&dist[j]>mat[u][j])dist[j]=mat[u][j];
}
printf("%d\n",sum);
}
return 0;
}
還是暢通工程
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 29993 Accepted Submission(s): 13396
Problem Description 某省調查鄉村交通狀況，得到的統計表中列出了任意兩村莊間的距離。省政府“暢通工程”的目標是使全省任何兩個村莊間都可以實現公路交通（但不一定有直接的公路相連，只要能間接通過公路可達即可），並要求鋪設的公路總長度為最小。請計算最小的公路總長度。
Input 測試輸入包含若干測試用例。每個測試用例的第1行給出村莊數目N ( < 100 )；隨後的N(N-1)/2行對應村莊間的距離，每行給出一對正整數，分別是兩個村莊的編號，以及此兩村莊間的距離。為簡單起見，村莊從1到N編號。
當N為0時，輸入結束，該用例不被處理。
Output 對每個測試用例，在1行裡輸出最小的公路總長度。
Sample Input
3
1 2 1
1 3 2
2 3 4
4
1 2 1
1 3 4
1 4 1
2 3 3
2 4 2
3 4 5
0
Sample Output
3
5
HintHint
Huge input, scanf is recommended.
Source 浙大計算機研究生復試上機考試-2006年 代碼：
#include
#include
#define MAX 99999999
int mat[105][105];
int main()
{
int n,i,j,k,sum,vis[1005],a,b,c,dist[105];
while(scanf("%d",&n)!=EOF)
{
if(n==0)break;
memset(vis,0,sizeof(vis));
for(i=0;i<105;i++)
for(j=0;j<105;j++)
mat[i][j]=MAX;
for(i=1;i<=n*(n-1)/2;i++)
{
scanf("%d%d%d",&a,&b,&c);
mat[a][b]=mat[b][a]=c;
}
vis[1]=1;
dist[1]=0;
sum=0;
int pos=1;
for(i=2;i<=n;i++)
{
dist[i]=mat[1][i];
}
for(i=1;i {
int mini=MAX,u=-1;
for(j=1;j<=n;j++)
if(!vis[j]&&dist[j] {
u=j;mini=dist[j];
}
vis[u]=1;
sum=sum+dist[u];
for(j=1;j<=n;j++)
if(!vis[j]&&dist[j]>mat[u][j])dist[j]=mat[u][j];
}
printf("%d\n",sum);
}
return 0;
}
兩題都是最簡單的最小生成樹prime算法的套用。。。。
