poj1258 最小生成樹

Agri-Net

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 56761		Accepted: 23531

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

最小生成樹裸題，用prim或kruskal算法。

prim：

#include<iostream>
#include<cmath>
using namespace std;
int n;///點的數量
int mapp[101][101];///存鄰接表
int dis[101];///表示每個點到生成樹的距離
int inf=0xfffffff;
int min_node;///距離生成樹最近的點
int min_arc;///距離生成樹最近的點與生成樹的距離（權值）
int sum;///最小生成樹權值和

int min(int a,int b)
{
    return a<b?a:b;
}
void prim()
{
    int now;///表示符合加入生成樹的點
    sum=0;
    for(int i=1;i<=n;i++)
        dis[i]=inf;
    now=1;///第一個點是起始點
    for(int i=1;i<n;i++)
    {
        dis[now]=-1;///將now點加入生成樹
        min_arc=inf;
        for(int j=1;j<=n;j++)///把剩下的點到生成樹的距離刷新一遍
        {
            if(now!=j && dis[j]>=0)///判斷是不是生成樹中的點
            {
                dis[j]=min(dis[j],mapp[now][j]);
                if(dis[j] < min_arc)///找到下一個加入生成樹的點
                {
                    min_arc=dis[j];
                    min_node=j;
                }
            }
        }
        now=min_node;
        sum+=min_arc;
    }
}

int main()
{
    while(cin>>n){
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
        cin>>mapp[i][j];
    prim();
    cout<<sum<<endl;}
    return 0;
}

kruskal：

#include<iostream>
#include<algorithm>
using namespace std;

struct edge
{
    int u,v,w;
};///定義邊
int p[110];///並查集
edge e[110*110];
int n,ne;///節點數,邊數

int cmp(edge &a,edge &b)
{
    return a.w < b.w;
}

int find(int i)///返回p[i]所在樹的樹根
{
    return p[i]==i?i:p[i]=find(p[i]);
}

bool union_set(int a,int b)
{
    a=find(a);
    b=find(b);
    if(a!=b)///沒有共同祖先
    {
        p[a]=b;///將節點納入最小生成樹集合中
        return true;
    }
    else
        return false;
}
int kruskal()
{
    int mst_edge=0,sum=0;
    for(int i=1;i<=n;i++)///初始化並查集
        p[i]=i;
    sort(e,e+ne,cmp);///將邊按升序排序
    for(int i=0;i<ne;i++)
    {
        ///如果加入的邊不會使樹形成迴路
        if(union_set(e[i].u,e[i].v))
        {
            sum+=e[i].w;
            if(++mst_edge==n-1)
                return sum;
        }
    }
    return mst_edge;
}
int main()
{
    int cost;
    while(cin>>n)
    {
        ne=0;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                cin>>cost;
                if(i!=j)
                {
                    e[ne].u=i;
                    e[ne].v=j;
                    e[ne++].w=cost;
                }
            }
        }
        cout<<kruskal()<<endl;
    }
    return 0;
}