[ACM] poj 1258 Agri-Net (最小生成树）

Agri-Net

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 37131		Accepted: 14998

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 <iostream>
#include <stdio.h>
#include <algorithm>
using namespace std;
const int maxn=5010;
int parent[110];
int n;

struct Node
{
    int from,to,edge;
}node[maxn];

void init(int n)
{
    for(int i=1;i<=n;i++)
        parent[i]=i;
}

int find(int x)
{
    return parent[x]==x?x:find(parent[x]);
}

bool cmp(Node a,Node b)
{
    if(a.edge<b.edge)
        return true;
    return false;
}

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        int m=1;
        int len;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
            {
                scanf("%d",&len);
                if(i<j)
                {
                    int temp=m;
                    node[temp].from=i;
                    node[temp].to=j;
                    node[temp].edge=len;
                    m++;
                }
            }
        init(n);
        m=n*(n-1)/2;
        len=0;
        sort(node+1,node+1+m,cmp);
        for(int i=1;i<=m;i++)
        {
            int x=find(node[i].from);
            int y=find(node[i].to);
            if(x==y)
                continue;
            else
            {
                len+=node[i].edge;
                parent[x]=y;
            }
        }
        printf("%d\n",len);
    }
    return 0;
}

prim算法：

代码：

#include <stdio.h>
#include <iostream>
#include <string.h>
using namespace std;
const int maxn=110;
const int inf=0x7fffffff;
int map[maxn][maxn],low[maxn],visit[maxn];
//map数组用来记录地图，low数组用来保存与已增加树中的顶点相连的边。（保证权值最小的肯定在里边）。visit用来记录该顶点是否已增加树中
int n;

int prim()
{
    int pos,Min,result=0;
    memset(visit,0,sizeof(visit));
    visit[1]=1,pos=1;//首先找的是1这个顶点，增加图中
    for(int i=1;i<=n;i++)
        if(i!=pos)
        low[i]=map[pos][i];//与1顶点相连的边地权值
    for(int i=1;i<n;i++)//每次增加一条边。n个顶点增加n-1条边。循环n-1次就能够了
    {
        Min=inf;
        for(int j=1;j<=n;j++)
            if(!visit[j]&&Min>low[j])
            {
                Min=low[j];//Min找到的是与已增加图中的顶点相连的边的最小值
                pos=j;//该顶点为j
            }
        result+=Min;//加上最小边
        visit[pos]=1;//新的顶点位置
        for(int j=1;j<=n;j++)
            if(!visit[j]&&low[j]>map[pos][j])
            low[j]=map[pos][j];//更新low[]值，新的与图中已有的点相连的边，假设比原来小的，就更新
    }
    return result;
}

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
            {
                scanf("%d",&map[i][j]);
                map[j][i]=map[i][j];
            }
        //对于题目中输入了map[i][i]的值的话。map数组不用预处理，假设没输入,那么memset(map,0x3f3f3f3f，sizeof(map))；最大化处理
        cout<<prim()<<endl;
    }
    return 0;
}