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hdu 1599find the mincost route(Floyd算法,最小环无向图)

2019年11月16日 - 计算机教程

hdu 1599find the mincost route(Floyd算法,最小环无向图)

poj1258 Agri-Net +hdu 1233 还是畅通工程 (最小生成树Prime算法)

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

代码:
#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[402.com,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;
}

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