标签：acm poj 最小生成树
Agri-Net
| Time Limit: 1000MS |
|
Memory Limit: 10000K |
| Total Submissions: 39499 |
|
Accepted: 16017 |
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
最小生成树的入门题；水题一道；帮助理解一下prim算法；prim算法刚开始看的时候，没有怎么搞懂，最近又回过头来看最小生成树这一块，感觉对prim算法的理解就更深刻了；最小生成树是包含n个点的n-1条边的权值最小的一个子图；prim算法是从一个点开始，寻找与它相连的权值最小的一个点，然后把这个点也加入进来，比较这与这两个点相连的权值最小的点，更新之间的权值，后面的点以此类推；用点来寻找边；直到包含有n个点。
下面是代码；
用一个二维数组map储存图结构，用low数组记录两点之间的权值，用visit数组标志这个店是否访问；
#include <cstdio>
#include <cstring>
const int Max=0x3f3f3f3f;
const int maxn=100+10;
int map[maxn][maxn],low[maxn],visit[maxn];
int n;
int prim()
{
int pos,i,j,min,sum=0;
memset(visit,0,sizeof(visit));//初始化visit数组
visit[1]=1;//从第一个点开始
pos=1;//标记和记录这个点
for(i=1;i<=n;i++)
low[i]=map[pos][i];//用low数组记录权值
for(i=1;i<n;i++) //第一个点已经进行了，还需要进行n-1次；
{
min=Max;//把min赋初值
for(j=1;j<=n;j++)
{
if(visit[j]==0&&low[j]!=0&&low[j]<min)//比较权值的大小
{
min=low[j];
pos=j;//记录权值最小的点，下一次从这个点开始
}
}
sum+=min;//记录权值的和
visit[pos]=1;//标记访问
for(j=1;j<=n;j++)//访问下一个点
{
if(visit[j]==0&&low[j]>map[pos][j])
low[j]=map[pos][j];
}
}
return sum;
}
int main()
{
int sum,i,j;
while(scanf("%d",&n)!=EOF)
{
sum=0;
memset(map,Max,sizeof(map));
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
scanf("%d",&map[i][j]);
sum=prim();
printf("%d\n",sum);
}
return 0;
}
poj 1258 Agri-Net （最小生成树 prim）,布布扣,bubuko.com
poj 1258 Agri-Net （最小生成树 prim）
标签：acm poj 最小生成树
原文地址：http://blog.csdn.net/whjkm/article/details/38271731
