最小生成树 简单模板题 POJ1251&&POJ1258

Description

The Head Elder of the tropical island of Lagrishan has a problem. A burst of foreign aid money was spent on extra roads between villages some years ago. But the jungle overtakes roads relentlessly, so the large road network is too expensive to maintain. The Council of Elders must choose to stop maintaining some roads. The map above on the left shows all the roads in use now and the cost in aacms per month to maintain them. Of course there needs to be some way to get between all the villages on maintained roads, even if the route is not as short as before. The Chief Elder would like to tell the Council of Elders what would be the smallest amount they could spend in aacms per month to maintain roads that would connect all the villages. The villages are labeled A through I in the maps above. The map on the right shows the roads that could be maintained most cheaply, for 216 aacms per month. Your task is to write a program that will solve such problems. 

Input

The input consists of one to 100 data sets, followed by a final line containing only 0. Each data set starts with a line containing only a number n, which is the number of villages, 1 < n < 27, and the villages are labeled with the first n letters of the alphabet, capitalized. Each data set is completed with n-1 lines that start with village labels in alphabetical order. There is no line for the last village. Each line for a village starts with the village label followed by a number, k, of roads from this village to villages with labels later in the alphabet. If k is greater than 0, the line continues with data for each of the k roads. The data for each road is the village label for the other end of the road followed by the monthly maintenance cost in aacms for the road. Maintenance costs will be positive integers less than 100. All data fields in the row are separated by single blanks. The road network will always allow travel between all the villages. The network will never have more than 75 roads. No village will have more than 15 roads going to other villages (before or after in the alphabet). In the sample input below, the first data set goes with the map above. 

Output

The output is one integer per line for each data set: the minimum cost in aacms per month to maintain a road system that connect all the villages. Caution: A brute force solution that examines every possible set of roads will not finish within the one minute time limit. 

Sample Input

9
A 2 B 12 I 25
B 3 C 10 H 40 I 8
C 2 D 18 G 55
D 1 E 44
E 2 F 60 G 38
F 0
G 1 H 35
H 1 I 35
3
A 2 B 10 C 40
B 1 C 20
0

Sample Output

216
30
裸的
#define inf 0x3f3f3f3f
#include <iostream>
#include <string.h>
using namespace std;
#define Max 30
int map[Max][Max];
int dis[Max];
int vis[Max];
int n;
int prim()
{
    int count=0;
    int minncost;
    memset(vis,0,sizeof(vis));
    vis[1]=1;
    int pos=1;
    for(int i=1; i<=n; i++)
    {
        if(i!=pos)
            dis[i]=map[pos][i];
    }
    for(int i=1; i<n; i++)
    {
        minncost=inf;
        for(int j=1; j<=n; j++)
        {
            if(!vis[j]&&dis[j]<minncost)
            {
                minncost=dis[j];
                pos=j;
            }
        }
        count+=minncost;
        vis[pos]=1;
        for(int j=1; j<=n; j++)
        {
            if(!vis[j]&&dis[j]>map[pos][j])
                dis[j]=map[pos][j];
        }

    }
    return count;

}
int main()
{
    while(cin>>n&&n)
    {
        memset(map,inf,sizeof(map));
        for(int i=0; i<n-1; i++)
        {
            char x,y;
            int m;
            cin>>x>>m;
            for(int j=1; j<=m; j++)
            {
                int z;
                cin>>y>>z;
                map[x-64][y-64]=map[y-64][x-64]=z;
            }
        }
        cout<<prim()<<endl;
    }
    return 0;
}

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
代码几乎一样
#include<stdio.h>
#include<string.h>
int map[110][110];
int vis[110];
int dis[110];
#define inf 0x3f3f3f3f
int n;
int prim()
{
    int count=0;
    int minncost;
    memset(vis,0,sizeof(vis));
    vis[1]=1;
    int pos=1;
    for(int i=1; i<=n; i++)
    {
        if(i!=pos)
            dis[i]=map[pos][i];
    }
    for(int i=1; i<n; i++)
    {
        minncost=inf;
        for(int j=1; j<=n; j++)
        {
            if(!vis[j]&&dis[j]<minncost)
            {
                minncost=dis[j];
                pos=j;
            }
        }
        count+=minncost;
        vis[pos]=1;
        for(int j=1; j<=n; j++)
        {
            if(!vis[j]&&dis[j]>map[pos][j])
                dis[j]=map[pos][j];
        }

    }
    return count;

}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        int c;
        memset(map,inf,sizeof(map));
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
            {
                scanf("%d",&c);
                map[i][j]=c;
            }
        int ans=prim();
        printf("%d\n",ans);
    }
}