POJ 1251 Jungle Roads (最小生成树prim)

本文介绍了一个关于最小生成树的实际问题及其解决方案。问题背景设定在一个热带岛屿上,需要选择最经济的道路维护方案来连接各个村庄。通过使用Prim算法,文章提供了一种高效的方法来找出连接所有村庄所需的最低维护成本。


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
题解:

  很水的题,直接套模板。需要注意的是输入格式,因为每行末尾可能会有多个空格。用cin输入就不用担心。
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
using namespace std;
const int maxn=105,INF=1e9;
int cost[maxn][maxn];
int mincost[maxn];
bool used[maxn];
int n;
int prim()
{
    for(int i=0;i<n;i++)
    {
        mincost[i]=INF;
        used[i]=false;
    }
    mincost[0]=0;
    int res=0;
    while(666)
    {
        int v=-1;
        for(int u=0;u<n;u++)
        {
            if(!used[u]&&(v==-1||mincost[u]<mincost[v]))
                v=u;
        }
        if(v==-1)
            break;
        used[v]=true;
        res+=mincost[v];
        for(int u=0;u<n;u++)
            mincost[u]=min(mincost[u],cost[v][u]);
    }
    return res;
}
int main()
{
    while(cin>>n,n)
    {
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
                cost[i][j]=INF;
        for(int i=0;i<n-1;i++)
        {    
            char u;
            int k;
            cin>>u>>k;
            while(k--)
            {
                char v;
                int val;
                cin>>v>>val;
                cost[u-'A'][v-'A']=cost[v-'A'][u-'A']=val;
            }
        }    
        cout<<prim()<<endl;
    }
    return 0;
}

 

 

转载于:https://www.cnblogs.com/orion7/p/7373756.html

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值